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((-1+x)/x)^(5*x)
Limit of the function/ ((-1+x)/x)^(5*x)

Limit of the function ((-1+x)/x)^(5*x)

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You have entered [src] 
             5*x
     /-1 + x\   
 lim |------|   
x->oo\  x   /
lim⁡x→∞(x−1x)5x\lim_{x \to \infty} \left(\frac{x - 1}{x}\right)^{5 x} 
Limit(((-1 + x)/x)^(5*x), x, oo, dir='-')
Detail solution
Let's take the limit
lim⁡x→∞(x−1x)5x\lim_{x \to \infty} \left(\frac{x - 1}{x}\right)^{5 x}
transform
lim⁡x→∞(x−1x)5x\lim_{x \to \infty} \left(\frac{x - 1}{x}\right)^{5 x}
=
lim⁡x→∞(x−1x)5x\lim_{x \to \infty} \left(\frac{x - 1}{x}\right)^{5 x}
=
lim⁡x→∞(−1x+xx)5x\lim_{x \to \infty} \left(- \frac{1}{x} + \frac{x}{x}\right)^{5 x}
=
lim⁡x→∞(1−1x)5x\lim_{x \to \infty} \left(1 - \frac{1}{x}\right)^{5 x}
=
do replacement
u=x−1u = \frac{x}{-1}
then
lim⁡x→∞(1−1x)5x\lim_{x \to \infty} \left(1 - \frac{1}{x}\right)^{5 x} =
=
lim⁡u→∞(1+1u)−5u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 5 u}
=
lim⁡u→∞(1+1u)−5u\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 5 u}
=
((lim⁡u→∞(1+1u)u))−5\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-5}
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))−5=e−5\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-5} = e^{-5}

The final answer:
lim⁡x→∞(x−1x)5x=e−5\lim_{x \to \infty} \left(\frac{x - 1}{x}\right)^{5 x} = e^{-5}
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
02468-8-6-4-2-10100.00.5
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Rapid solution [src] 
 -5
e
e−5e^{-5}
Expand and simplify
The graph
Limit of the function ((-1+x)/x)^(5*x)
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