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Find the limit of $\frac{\sin\left(e^{-x}\right)}{x^{-1}}$ as $x$ approaches $\infty $

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Final Answer

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Step-by-step Solution

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The limit of the product of two functions is equal to the product of the limits of each function

$\lim_{x\to\infty }\left(\frac{1}{x^{-1}}\right)\lim_{x\to\infty }\left(\sin\left(e^{-x}\right)\right)$

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$\lim_{x\to\infty }\left(\frac{1}{x^{-1}}\right)\lim_{x\to\infty }\left(\sin\left(e^{-x}\right)\right)$

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Learn how to solve problems step by step online. Find the limit of sin(e^(-x))/(x^(-1)) as x approaches infinity. The limit of the product of two functions is equal to the product of the limits of each function. Because sine is a continuous function, we can bring the limit inside of the sine. Apply the power rule of limits: \displaystyle{\lim_{x\to a}f(x)^{g(x)} = \lim_{x\to a}f(x)^{\displaystyle\lim_{x\to a}g(x)}}. The limit of a constant is just the constant.

Final Answer

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Limits by Direct SubstitutionLimits by L'Hôpital's ruleLimits by FactoringLimits by Rationalizing

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1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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