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

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$\frac{1}{e}$$\,\,\left(\approx 0.36787944117144233\right) Got another answer? Verify it here ## Step-by-step Solution Problem to solve: \lim_{x\to\infty}\left(1-\frac{1}{x}\right)^x Specify the solving method 1 Rewrite the limit using the identity: a^x=e^{x\ln\left(a\right)} \lim_{x\to\infty }\left(e^{x\ln\left(1+\frac{-1}{x}\right)}\right) Learn how to solve limits to infinity problems step by step online. \lim_{x\to\infty }\left(e^{x\ln\left(1+\frac{-1}{x}\right)}\right) Learn how to solve limits to infinity problems step by step online. Find the limit of (1-1/x)^x as x approaches \infty. Rewrite the limit using the identity: a^x=e^{x\ln\left(a\right)}. Apply the power rule of limits: \displaystyle{\lim_{x\to a}f(x)^{g(x)} = \lim_{x\to a}f(x)^{\displaystyle\lim_{x\to a}g(x)}}. The limit of a constant is just the constant. Rewrite the product inside the limit as a fraction. ## Final Answer \frac{1}{e}$$\,\,\left(\approx 0.36787944117144233\right)$
SnapXam A2

### beta Got another answer? Verify it!

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0
a
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g
m
n
u
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w
x
y
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.
(◻)
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◻/◻
/
÷
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

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