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Resoluo Limites
a)
b)
c)
d)
e)
f)
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g)
h)
i)
j)
LIMITES DE FUNES QUANDO X SE APROXIMA DE UMA CONSTANTE
.
O seguintes problemas requerem o uso do clculo de limites de funes ao x se aproximar deuma constante. A maioria dos problemas de nvel mdio. Alguns so engenhosos. Todas as soluesso dadas sem o uso da regra de L'Hopital
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PROBLEMA 1 :Calcule
.
o PROBLEMA 2 :Calcule
.
o PROBLEMA 3 :Calcule
.
o PROBLEMA 4 :Calcule
.
o PROBLEMA 5 :Calcule
.
o PROBLEMA 6 :Calcule
.
o PROBLEMA 7 :Calcule
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.
o PROBLEMA 8 :Calcule
.
o PROBLEMA 9 :Calcule
.
o PROBLEMA 10 :Calcule
o .o PROBLEMA 11 :Calcule
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o PROBLEMA 12 :Calcule
.
o PROBLEMA 13 :Calcule
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.
Este PROBLEMA requer uma substituio especial, identidades trigonomtricas, elimites trigonomtricos.
O seguinte problema requer a noo de limites laterais
o PROBLEMA 14 :Considere a funo
i.) Desenhe o grfico de f.
ii.) Determine os sguintes limites.
a.)
b.)
c.)
d.) e.)
f.)
g.)
h.)
i.)
j.)
k.)
l.)o PROBLEMA 15 :Consider the function
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Determine os valores das constantes ae btal que existe e seja igual a f(2).
SOLUTIONS TO LIMITS OF FUNCTIONS AS X APPROACHES ACONSTANT
SOLUTION 1 :
.
ClickHERE to return to the list of problems.
SOLUTION 2 :
(Circumvent the indeterminate form by factoring both the numerator and denominator.)
(Divide out the factorsx - 2 , the factors which are causing the indeterminate form . Now
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%201http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2018/2/2019 Resoluo Limites
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the limit can be computed. )
ClickHERE to return to the list of problems.
SOLUTION 3 :
(Circumvent the indeterminate form by factoring both the numerator and denominator.)
(Divide out the factorsx - 3 , the factors which are causing the indeterminate form . Nowthe limit can be computed. )
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%202http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%202http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%202http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2028/2/2019 Resoluo Limites
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.
ClickHERE to return to the list of problems.
SOLUTION 4 :
(Algebraically simplify the fractions in the numerator using a common denominator.)
(Division by is the same as multiplication by .)
(Factor the denominator . Recall that .)
(Divide out the factorsx + 2 , the factors which are causing the indeterminate form . Nowthe limit can be computed. )
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%203http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%203http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%203http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2038/2/2019 Resoluo Limites
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.
ClickHERE to return to the list of problems.
SOLUTION 5 :
(Eliminate the square root term by multiplying by the conjugate of the numerator over itself.Recall that
. )
(Divide out the factorsx - 4 , the factors which are causing the indeterminate form . Nowthe limit can be computed. )
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%204http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%204http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%204http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2048/2/2019 Resoluo Limites
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.
ClickHERE to return to the list of problems.
SOLUTION 6 :
(It may appear that multiplying by the conjugate of the numerator over itself is a reasonablenext step.
It's a good idea, but doesn't work. Instead, writex - 27 as the difference of cubes and recall that
.)
(Divide out the factors , the factors which are causing the indeterminate form .Now the limit can be computed. )
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%205http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%205http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%205http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2058/2/2019 Resoluo Limites
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= 27 .
ClickHERE to return to the list of problems.
SOLUTION 7 :
(Multiplying by conjugates won't work for this challenging problem. Instead, recall that
and , and
note that and . This should help explain the nextfew mysterious steps.)
(Divide out the factorsx - 1 , the factors which are causing the indeterminate form . Nowthe limit can be computed. )
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%206http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%206http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%206http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2068/2/2019 Resoluo Limites
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.
ClickHERE to return to the list of problems.
SOLUTION 8 :
(If you wrote that , you are incorrect. Instead, multiply and divide by 5.)
(Use the well-known fact that .)
.
ClickHERE to return to the list of problems.
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%207http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%207http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%207http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%208http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%208http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%208http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%208http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2078/2/2019 Resoluo Limites
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SOLUTION 9 :
(Recall the trigonometry identity .)
(The numerator is the difference of squares. Factor it.)
(Divide out the factors , the factors which are causing the indeterminate form .Now the limit can be computed. )
.
ClickHERE to return to the list of problems.
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%209http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%209http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%209http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2098/2/2019 Resoluo Limites
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SOLUTION 10 :
(Factorx from the numerator and denominator, then divide these factors out.)
(The numerator approaches -7 and the denominator is a positve quantity approaching 0 .)
(This is NOT an indeterminate form. The answer follows.)
.
(Thus, the limit does not exist.)
ClickHERE to return to the list of problems.
SOLUTION 11 :
(The numerator approaches -3 and the denominator is a negative quantity which approaches 0 asx
approaches 0 .)
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2010http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2010http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2010http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%20108/2/2019 Resoluo Limites
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(This is NOT an indeterminate form. The answer follows.)
.
(Thus, the limit does not exist.)
ClickHERE to return to the list of problems.
SOLUTION 12 :
(Recall that . )
(Divide out the factorsx - 1 , the factors which are causing the indeterminate form . Nowthe limit can be computed. )
.
(The numerator approaches 1 and the denominator approaches 0 as x approaches 1 . However,
the quantity
in the denominator is sometimes negative and sometimes positive. Thus, the correct answer isNEITHER
NOR . The correct answer follows.)
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2011http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2011http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2011http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%20118/2/2019 Resoluo Limites
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The limit does not exist.
ClickHERE to return to the list of problems.
SOLUTION 13 :
(Make the replacement so that . Note that
asx approaches , h approaches 0 . )
(Recall the well-known, but seldom-used, trigonometry
identity .)
(Recall the well-known trigonometry identity . )
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2012http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2012http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2012http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%20128/2/2019 Resoluo Limites
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(Recall that . )
= 2 .
ClickHERE to return to the list of problems.
The next problem requires an understanding of one-sided limits.
SOLUTION 14 : Consider the function
i.) The graph offis given below.
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2013http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2013http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2013http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%20138/2/2019 Resoluo Limites
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ii.) Determine the following limits.
a.) .
b.) .
c.) We have that does not exist since does not equal .
d.) .
e.) .
f.) We have that since .
g.) We have that (The numerator is
always -1 and the denominator is always a positive number approaching 0.) , so the limit does
not exist.
h.) .
i.) We have that does not exist since does not equal .
j.) .
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k.) .
l.) .
ClickHERE to return to the list of problems.
SOLUTION 15 : Consider the function
Determine the values of constants a and b so that exists. Begin by computing one-
sided limits atx=2 and setting each equal to 3. Thus,
and
.
Now solve the system of equations
a+2b = 3 and b-4a = 3 .
Thus,
a = 3-2b so that b-4(3-2b) = 3
iffb-12+ 8b = 3
iff 9b = 15
iff .
Then
http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2014http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2014http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%2014http://mtm.ufsc.br/~azeredo/calculos/Acalculo/x/limits/LimitConstant.html#PROBLEM%20148/2/2019 Resoluo Limites
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LIMITES DE FUNES QUANDO X SE APROXIMA DO INFINITO
Os seguintes problemas requerem o clculo algbrico de limites de funes ao
x se aproximar de + ou - infinito. Alguns so engenhosos. Todas as solues soobtidas sem a regra de L'Hopital. Tente obter suas solues sem olhar as solues
dadas levando cuidadosamente em considerao as
formas e durante seus clculos. Inicialmente alguns estudantes
incorretamante concluem que igual a 1 , ou que o limite no existe, ou
ou . Muitos concluem que igual a 0. De fato, as
formas e so exemplos de formas indeterminadas . Isto significa que
voc ainda no determinou uma resposta. Usualmente estas formas indeterminadas
podem ser contornadas por manipulaes algbricas. As solues esto todas em
ingls.
o PROBLEMA 1 : Calcule
.
o
PROBLEMA 2 :
Calcule
.
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o PROBLEMA 3 : Calcule
.
o PROBLEMA 4 :
.
o PROBLEMA 5 :
.
o PROBLEMA 6 :
.
o PROBLEMA 7 : Calcule
.
o PROBLEMA 8 : Calcule
.
SOLUTIONS TO LIMITS OF FUNCTIONS AS X APPROACHES PLUS ORMINUS INFINITY
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SOLUTION 1 :
= = 0 .
(The numerator is always 100 and the denominator approaches as x approaches ,so that the resulting fraction approaches 0.)
ClickHERE to return to the list of problems.
SOLUTION 2 :
= = 0 .
(The numerator is always 7 and the denominator approaches as xapproaches , so that the resulting fraction approaches 0.)
ClickHERE to return to the list of problems.
SOLUTION 3 :
=
(This is NOT equal to 0. It is an indeterminate form. It can be circumvented by factoring.)
=
(As x approaches , each of the two expressions and 3x - 100 approaches .)
=
http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%201http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%201http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%201http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%202http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%202http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%202http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%202http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2018/2/2019 Resoluo Limites
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(This is NOT an indeterminate form. It has meaning.)
= .
(Thus, the limit does not exist. Note that an alternate solution follows by first factoringout , the highest power ofx . Try it.)
ClickHERE to return to the list of problems.
SOLUTION 4 :
=
(As x approaches , each of the two expressions and approaches . )
=
(This is NOT an indeterminate form. It has meaning.)
= .
(Thus, the limit does not exist. Note that an alternate solution follows by first factoring
out , the highest power ofx . Try it.)
ClickHERE to return to the list of problems.
SOLUTION 5 :
(Note that the expression leads to the indeterminate form . Circumventthis by appropriate factoring.)
= .
http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%203http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%203http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%203http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%204http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%204http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%204http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%204http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2038/2/2019 Resoluo Limites
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(As x approaches , each of the three expressions , , andx - 10 approaches .)
=
=
= .
(Thus, the limit does not exist. Note that an alternate solution follows by first factoringout , the highest power ofx . Try it. )
ClickHERE to return to the list of problems.
SOLUTION 6 :
=
(This is an indeterminate form. Circumvent it by dividing each term byx .)
=
=
=
http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%205http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%205http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%205http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2058/2/2019 Resoluo Limites
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(Asx approaches , each of the two expressions and approaches 0.)
=
= .
ClickHERE to return to the list of problems.
SOLUTION 7 :
(Note that the expression leads to the indeterminate form as xapproaches . Circumvent this by dividing each of the terms in the original problem by .)
=
=
=
http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%206http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%206http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%206http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2068/2/2019 Resoluo Limites
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(Each of the three expressions , , and approaches 0 asx approaches .)
=
= .
ClickHERE to return to the list of problems.
SOLUTION 8 :
(Note that the expression leads to the indeterminate
form asx approaches . Circumvent this by dividing each of the terms in theoriginal problem by , the highest power ofx in the problem . This is not the only step thatwill work here. Dividing by , the highest power ofx in the numerator, also leads to the
correct answer. You might want to try it both ways to convince yourself of this.)
=
=
http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%207http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%207http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%207http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2078/2/2019 Resoluo Limites
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=
(Each of the five expressions , , , , and approaches 0 asx approaches .)
=
= 0 .
ClickHERE to return to the list of problems.
SOLUTION 9 :
(Note that the expression leads to the indeterminate
form asx approaches . Circumvent this by dividing each of the terms in theoriginal problem by , the highest power ofx in the problem. . This is not the only step thatwill work here. Dividing byx , the highest power ofx in the denominator, actually leads moreeasily to the correct answer. You might want to try it both ways to convince yourself of this.)
=
=
http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%208http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%208http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%208http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2088/2/2019 Resoluo Limites
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=
(Each of the three expressions , , and approaches 0 asx approaches .)
=
=
(This is NOT an indeterminate form. It has meaning. However, to determine it's exactmeaning requires a bit more analysis of the origin of the 0 in the denominator. Note
that = . It follows that ifx is a negative number then both of the
expressions and are negative so that is positive. Thus, for the
expression the numerator approaches 7 and the denominator is a positivequantity approaching 0 asx approaches . The resulting limit is .)
= .
(Thus, the limit does not exist.)
ClickHERE to return to the list of problems.
SOLUTION 10 :
http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%209http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%209http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%209http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2098/2/2019 Resoluo Limites
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=
(You will learn later that the previous step is valid because of the continuity of the square rootfunction.)
=
(Inside the square root sign lies an indeterminate form. Circumvent it by dividing each termby , the highest power ofxinside the square root sign.)
=
=
=
(Each of the two expressions and approaches 0 asx approaches .)
=
=
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= .
ClickHERE to return to the list of problems.
SOLUTION 11 :
= `` ''
(Circumvent this indeterminate form by using the conjugate of the
expression in an appropriate fashion.)
=
(Recall that .)
=
=
=
=
= 0 .
ClickHERE to return to the list of problems.
http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2010http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2010http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2010http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2011http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2011http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2011http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%2011http://jurere.mtm.ufsc.br/~azeredo/x/limitinf/LimitInfinity.html#PROBLEM%20108/2/2019 Resoluo Limites
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SOLUTION 12 :
=
(This is NOT an indeterminate form. It has meaning.)
= .
(Thus, the limit does not exist.)
CONTINUIDADE DE FUNES DE UMA VARVEL
Os seguintes problemas envolvem a continuidade de funes de uma varivel.Umafunoy =f(x) contnua em um pontox=ase as seguintes trs condies soverseicadas:
i.)f(a) est definida,
ii.) exte (i.e., finito) ,
e
iii.) .
Uma funof dita ser contnua no intervaloIsef contnua em cadapontox emI. Aqui est uma lista de alguns fatos bem conhecidos relacionados acontinuidade:
1. A SOMA de funes contnuas contnua.
2. A DIFERENA de funes contnuas contnua.
3. O PRODUTO de funes contnuas contnua.
4. O QUOCIENTE de funes contnuas contnua em todos os
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pontosx aonde o DENOMINADOR NO ZERO.
5. A COMPOSIO de funes contnuas contnua em todos ospontosx onde a composio est propriamente definida.
6. Qualquer polinmio contnuo para todos os valores dex.
7. A funo ex e as funes trigonomtricas e so contnua paratodos os valores dex.
o PROBLEMA 1 : Determine se a seguinte funo contnua emx=1 .
o PROBLEMA 2 : Determine se a seguinte funo contnua emx=-2 .
o PROBLEMA 3 : Determine se a seguinte funo contnua emx=0 .
o PROBLEMA 4 : Determine se a funo contnua atx=-1 .
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o PROBLEMA 5 : Check the following function for continuity atx=3 andx=-3 .
o PROBLEMA 6 : Para que valores dexa funo contnua ?
o PROBLEMA 7 : Para que valores dexa funo
contnua ?
o
PROBLEMA 8 :
Para que valores dex
a funo contnua ?
o PROBLEMA 9 : Para que valores dexa funo contnua ?
o PROBLEMA 10 : Para que valores dexa funo contnua?
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o PROBLEMA 11 : Para que valores dex aseguinte funo contnua ?
o PROBLEMA 12 : Determine todos os valores da constanteA para que a seguinte
funo seja contnua para todos os valores dex.
o PROBLEMA 13 : Determine todos os va;ores das constantes A e B para que a funo
seja contnua para todos os valores dex.
o PROBLEMA 14 : Mostre que a seguinte funo contnua para todos os valores
dex.
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o PROBLEMA 15 : Seja
Mostre quef contnua para todos os valores dex . Mostre quefdiferencivel para todos os valores dex, mas que a derivada ,f' , NO contnua emx=0 .
o PROBLEMA 9 : Calcule
.
o PROBLEMA 10 : Calcule
.
o PROBLEMA 11 : Calcule
.
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o PROBLEMA 12 : Calcule
.
o PROBLEMA 13 : Calcule
.
o PROBLEMA 14 : Calcule
.
o PROBLEMA 15 : Calcule
.
o PROBLEMA 16 : Calcule
.
o PROBLEMA 17 : Calcule
.
o PROBLEMA 18 : Calcule
.
o PROBLEMA 19 : Calcule
.
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ii.) .
But
iii.) ,
so condition iii.) is not satisfied and functionfis NOT continuous atx=1 .
ClickHERE to return to the list of problems.
SOLUTION 2 : Functionfis defined atx=-2 since
i.)f(-2) = (-2)2 + 2(-2) = 4-4 = 0 .
The left-hand limit
= (-2)2 + 2(-2)
= 4 - 4
= 0 .
The right-hand limit
= (-2)3 - 6(-2)
= -8 + 12
= 4 .
Since the left- and right-hand limits are not equal, ,
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ii.) does not exist,
and condition ii.) is not satisfied. Thus, functionfis NOT continuous atx=-2 .
ClickHERE to return to the list of problems.
SOLUTION 3 : Functionfis defined atx=0 since
i.)f(0) = 2 .
The left-hand limit
= 2 .
The right-hand limit
= 2 .
Thus, exists with
ii.) .
Since
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iii.) ,
all three conditions are satisfied, andfis continuous atx=0 .
ClickHERE to return to the list of problems.
SOLUTION 4 : Function h is not defined atx=-1 since it leads to division by zero.Thus,
i.) h(-1)
does not exist, condition i.) is violated, and function h is NOT continuous atx = -1 .
ClickHERE to return to the list of problems.
SOLUTION 5 : First, check for continuity atx=3 . Functionfis defined atx=3 since
i.) .
The limit
(Circumvent this indeterminate form by factoring the numerator and thedenominator.)
(Recall thatA2 -B2 = (A-B)(A+B) andA3 -B3 = (A-B)(A2+AB+B2 ) . )
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(Divide out a factor of (x-3) . )
=
,
i.e.,
ii.) .
Since,
iii.) ,
all three conditions are satisfied, andfis continuous atx=3 . Now, check forcontinuity atx=-3 . Functionfis not defined atx = -3 because of division by zero.Thus,
i.)f(-3)
does not exist, condition i.) is violated, andfis NOT continuous atx=-3 .
ClickHERE to return to the list of problems.
SOLUTION 6 : Functionsy =x2 + 3x + 5 andy =x2 + 3x - 4 are continuous for allvalues ofx since both are polynomials. Thus, the quotient of these two
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functions, , is continuous for all values ofx where thedenominator,y=x2 + 3x - 4 = (x-1)(x+4) , does NOT equal zero. Since (x-1)(x+4) = 0forx=1 andx=-4 , functionfis continuous for all values ofx EXCEPTx=1 andx=-4 .
ClickHERE to return to the list of problems.
SOLUTION 7 : First describe function g using functional composition. Letf(x)
=x1/3 , , and k(x) =x20 + 5 . Function kis continuous for all valuesofx since it is a polynomial, and functionsfand h are well-known to be continuousfor all values ofx . Thus, the functional compositions
and
are continuous for all values ofx . Since
,
function g is continuous for all values ofx .
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SOLUTION 8 : First describe functionfusing functional composition. Let g(x) =x2 -
2x and . Function g is continuous for all values ofx since it is a
polynomial, and function h is well-known to be continuous for . Since g(x)
=x2 - 2x =x(x-2) , it follows easily that for and . Thus, thefunctional composition
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is continuous for and . Since
,
functionfis continuous for and .
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SOLUTION 9 : First describe functionfusing functional composition.
Let and . Since g is the quotient of polynomialsy =x-1andy =x+2 , function g is continuous for all values ofx EXCEPT wherex+2 = 0 ,i.e., EXCEPT forx = -2 . Function h is well-known to be continuous forx > 0 .
Since , it follows easily that g(x) > 0 forx < -2 andx > 1 . Thus, thefunctional composition
is continuous forx < -2 andx > 1 . Since
,
functionfis continuous forx < -2 andx > 1 .
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SOLUTION 10 : First describe functionfusing functional composition.
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Let and h(x) = ex, both of which are well-known to be continuous for all
values ofx . Thus, the numerator is continuous (the functionalcomposition of continuous functions) for all values ofx . Now consider the
denominator . Let g(x) = 4 ,h(x) =x2 - 9 , and .Functions g and h are continuous for all values ofx since both are polynomials, and it
is well-known that function kis continuous for . Since h(x) =x2 - 9 = (x-
3)(x+3) = 0 whenx=3 orx=-3 , it follows easily that for and ,
so that is continuous (the functional composition of
continuous functions) for and . Thus, the
denominator is continuous (the difference of continuous functions)
for and . There is one other important consideration. We must insurethat the DENOMINATOR IS NEVER ZERO. If
then
.
Squaring both sides, we get
16 =x2 - 9
so that
x2 = 25
when
x = 5 orx = -5 .
Thus, the denominator is zero ifx = 5 orx = -5 . Summarizing, the quotient of these
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continuous functions, , is continuous for and , butNOT forx = 5 andx = -5 .
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SOLUTION 11 : Consider separately the three component functions which
determinef. Function is continuous forx > 1 since it is the quotient ofcontinuous functions and the denominator is never zero. Functiony = 5 -3x is
continuous for since it is a polynomial. Function is continuousforx < -2 since it is the quotient of continuous functions and the denominator isnever zero. Now check for continuity offwhere the three components are joinedtogether, i.e., check for continuity atx=1 andx=-2 . Forx = 1 functionfis definedsince
i.)f(1) = 5 - 3(1) = 2 .
The right-hand limit
=
(Circumvent this indeterminate form one of two ways. Either factor the numerator asthe difference of squares, or multiply by the conjugate of the denominator overitself.)
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= 2 .
The left-hand limit
=
= 5 - 3(1)
= 2 .
Thus,
ii.) .
Since
iii.) ,
all three conditions are satisfied, and functionfis continuous atx=1 . Now check forcontinuity atx=-2 . Functionfis defined atx=-2 since
i.)f(-2) = 5 - 3(-2) = 11 .
The right-hand limit
=
= 5 - 3( -2)
= 11 .
The left-hand limit
=
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= -1 .
Since the left- and right-hand limits are different,
ii.) does NOT exist,
condition ii.) is violated, and functionfis NOT continuous atx=-2 . Summarizing,
functionfis continuous for all values ofxEXCEPTx=-2 .
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SOLUTION 12 : First, consider separately the two components which determine
functionf. Functiony =A2x -A is continuous for for any value ofA since it isa polynomial. Functiony = 4 is continuous forx < 3 since it is a polynomial. NowdetermineA so that functionfis continuous atx=3 . Functionfmust be definedatx=3 , so
i.)f(3)=A2 (3) -A = 3A2 -A .
The right-hand limit
=
=A2 (3) -A
= 3A2 -A .
The left-hand limit
=
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= 4 .
For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,
ii.) ,
so that
3A2 -A - 4 = 0 .
Factoring, we get
(3A - 4)(A + 1) = 0
for
orA = -1 .
For either choice ofA ,
iii.) ,
all three conditions are satisfied, andfis continuous atx=3 . Therefore, functionfis
continuous for all values ofx if orA = -1 .
ClickHERE to return to the list of problems.
SOLUTION 13 : First, consider separately the three components which determine
functionf. Functiony =Ax -B is continuous for for any valuesofA andB since it is a polynomial. Functiony = 2x2 + 3Ax +B is continuous for
for any values ofA andB since it is a polynomial. Functiony = 4 iscontinuous forx > 1 since it is a polynomial. Now determineA andB so thatfunctionfis continuous atx=-1 andx=1 . First, consider continuity atx=-1 .Functionfmust be defined atx=-1 , so
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i.)f(-1)=A(-1) -B = -A -B .
The left-hand limit
=
=A (-1) -B
= -A -B .
The right-hand limit
=
= 2(-1)2 + 3A(-1) +B
= 2 - 3A +B .
For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,
ii.) ,
so that
2A - 2B = 2 ,
or
(Equation 1)
A -B = 1 .
Now consider continuity atx=1 . Functionfmust be defined atx=1 , so
i.)f(1)= 2(1)2 + 3A(1) +B = 2 + 3A +B .
The left-hand limit
=
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= 2(1)2 + 3A(1) +B
= 2 + 3A +B .
The right-hand limit
=
= 4 .
For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,
ii.) ,
or
(Equation 2)
3A +B = 2 .
Now solve Equations 1 and 2 simultaneously. Thus,
A -B = 1 and 3A +B = 2
are equivalent to
A =B + 1 and 3A +B = 2 .
Use the first equation to substitute into the second, getting
3 (B + 1 ) +B = 2 ,
3B + 3 +B = 2 ,
and
4B = -1 .
Thus,
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and
.
For this choice ofA andB it can easily be shown that
iii.)
and
iii.) ,
so that all three conditions are satisfied at bothx=1 andx=-1 , and functionfiscontinuous at bothx=1 andx=-1 . Therefore, functionfis continuous for all values
ofx if and .
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SOLUTION 14 : First describefusing functional composition. Let g(x) = -1/x2 and h(x) = ex . Function h is well-known to be continuous for all values ofx .Function g is the quotient of functions continuous for all values ofx , and is thereforecontinuous for all values ofx exceptx=0 , thatx which makes the denominator zero.Thus, for all values ofx exceptx=0 ,
f(x) = h ( g(x) ) = eg(x)
= e-1/x2
is a continuous function (the functional composition of continuous functions). Nowcheck for continuity offatx=0 . Functionfis defined atx=0 since
i.)f(0) = 0 .
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The limit
(The numerator approaches -1 and the denominator is a positive number approachingzero.)
,
so that
= 0 ,
i.e.,
ii.) .
Since
iii.) ,
all three conditions are satisfied, andfis continuous atx=0 . Thus,fis continuous forall values ofx .
ClickHERE to return to the list of problems.
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ii.) .
Since
iii.) ,
all three conditions are satisfied, andfis continuous atx=0 . Thus,fis continuous for
all values ofx . Now show thatfis differentiable for all values ofx . For we
can differentiatefusing the product rule and the chain rule. That is, for thederivative offis
.
Use the limit definition of the derivative to differentiatefatx=0 . Then
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.
Use the Squeeze Principle to evaluate this limit. For
.
If , then
.
If , then
.
In either case,
,
and it follows from the Squeeze Principle that
.
Thus,fis differentiable for all values ofx . Check to see iff' is continuous atx=0 .The functionf' is defined atx=0 since
i.)f'(0) = 0 .
However,
ii.)
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does not exist since the values of oscillate between -1 and +1asx approaches zero. Thus, condition ii.) is violated, and the derivative ,f' , is notcontinuous atx=0 .
NOTE : The continuity of functionffor all values ofx also follows from the factthatfis differentiable for all values ofx .
LIMITES DE FUNES VIA O PRINCPIO DO SANDUCHE
Os seguintes problemas envolvem o clculo de limites usando o princpio do
sanduche que est dado abaixo:
SQUEEZE PRINCIPLE : Suponha que as funesf, g , e h satisfaam
e
.
Ento
.
(NOTA : A quantidade A pode ser um nmero finito , , or . A quantidade L
pode ser um nmero finito, , or . O princpio do sanduche usado emproblemas de limites onde os procedimentos algbricos (fatorao, conjugao emanipulao algbrica etc.) no so efetivos. O uso do princpio do sanduche requeranlise acurada, habilidade algbrica e uso cuidadoso de desigualdades.
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o PROBLEMA 1 : Calcule
.
o PROBLEMA 2 : Calcule
.
o PROBLEMA 3 : Calcule
.
o PROBLEMA 4 : Calcule
.
o PROBLEMA 5 : Calcule
.
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o PROBLEMA 6 : Calcule
.
.
o PROBLEMA 7 : Calcule
.
.
o PROBLEMA 8 : Assuma que existe
e . Encontre .
.
o PROBLEMA 9 : Considere um crculo de raio 1 centrado na origem e um ngulo
de radians, , no diagrama.
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a.) Considerando as reas dos tringulo reto setor OAD, setor OAC, e tringuloreto OBC, conclua que
.
b.) Use parte a.) e o princpio do sanduche para mostrar que
o PROBLEMA 10 : Suponha que
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Mostre quef contnua emx=0
SOLUTIONS TO LIMITS USING THE SQUEEZE PRINCIPLE
SOLUTION 1 : First note that
because of the well-known properties of the sine function. Since we are computing the limitasx goes to infinity, it is reasonable to assume thatx > 0 . Thus,
.
Since
,
it follows from the Squeeze Principle that
.
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SOLUTION 2 : First note that
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because of the well-known properties of the cosine function. Now multiply by -1, reversingthe inequalities and getting
or
.
Next, add 2 to each component to get
.
Since we are computing the limit asx goes to infinity, it is reasonable to assume thatx + 3 >0. Thus,
.
Since
,
it follows from the Squeeze Principle that
.
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SOLUTION 3 : First note that
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because of the well-known properties of the cosine function, and therefore
.
Since we are computing the limit asx goes to infinity, it is reasonable to assume that 3 -2x < 0. Now divide each component by 3 - 2x, reversing the inequalities and getting
,
or
.
Since
,
it follows from the Squeeze Principle that
.
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SOLUTION 4 : Note that DOES NOT EXIST since values
of oscillate between -1 and +1 asx approaches 0 from the left. However, this does
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NOT necessarily mean that does not exist ! ? #. Indeed,x3 < 0 and
forx < 0. Multiply each component byx3, reversing the inequalities and getting
or
.
Since
,
it follows from the Squeeze Principle that
.
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SOLUTION 5 : First note that
,
so that
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= (does not exist).
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SOLUTION 6 : First note that
,
so that
,
,
and
.
Then
=
=
http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%205http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%205http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%205http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%2058/2/2019 Resoluo Limites
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or
.
Now divide byx2 + 1 and multiply byx2 , getting
.
Then
=
=
=
=
= 0 .
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Similarly,
= 0 .
It follows from the Squeeze Principle that
= 0 .
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SOLUTION 8 : Since
=
and
= ,
it follows from the Squeeze Principle that
,
that is,
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.
Thus,
.
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SOLUTION 9 : a.) First note that (See diagram below.)
area of triangle OAD < area of sector OAC < area of triangle OBC .
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The area of triangle OAD is
(base) (height) .
The area of sector OAC is
(area of circle) .
The area of triangle OBC is
(base) (height) .
It follows that
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or
.
b.) If , then and , so that dividing by results in
.
Taking reciprocals of these positive quantities gives
or
.
Since
,
it follows from the Squeeze Principle that
.
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http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%209http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%209http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%209http://www.mtm.ufsc.br/~azeredo/calculos/Acalculo/x/sanduiche/SqueezePrinciple.html#PROBLEM%2098/2/2019 Resoluo Limites
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SOLUTION 10 : Recall that functionfis continuous atx=0 if
i.)f(0) is defined ,
ii.) exists ,
and
iii.) .
First note that it is given that
i.)f(0) = 0 .
Use the Squeeze Principle to compute . For we know that
,
so that
.
Since
it follows from the Squeeze Principle that
ii.) .
Finally,
iii.) ,
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