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(1-7/x)^x
Limit of the function/ (1-7/x)^x

Limit of the function (1-7/x)^x

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            x
     /    7\ 
 lim |1 - -| 
x->oo\    x/
lim⁡x→∞(1−7x)x\lim_{x \to \infty} \left(1 - \frac{7}{x}\right)^{x} 
Limit((1 - 7/x)^x, x, oo, dir='-')
Detail solution
Let's take the limit
lim⁡x→∞(1−7x)x\lim_{x \to \infty} \left(1 - \frac{7}{x}\right)^{x}
transform
do replacement
u=x−7u = \frac{x}{-7}
then
lim⁡x→∞(1−7x)x\lim_{x \to \infty} \left(1 - \frac{7}{x}\right)^{x} =
=
lim⁡u→∞(1+1u)−7u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 7 u}
=
lim⁡u→∞(1+1u)−7u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 7 u}
=
((lim⁡u→∞(1+1u)u))−7\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-7}
The limit
lim⁡u→∞(1+1u)u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}
is second remarkable limit, is equal to e ~ 2.718281828459045
then
((lim⁡u→∞(1+1u)u))−7=e−7\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-7} = e^{-7}

The final answer:
lim⁡x→∞(1−7x)x=e−7\lim_{x \to \infty} \left(1 - \frac{7}{x}\right)^{x} = e^{-7}
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
02468-8-6-4-2-10100.01.0
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Rapid solution [src] 
 -7
e
e−7e^{-7}
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The graph
Limit of the function (1-7/x)^x
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