Mister Exam

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Limit of the function ((-2+x)/x)^(2*x)

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             2*x
     /-2 + x\   
 lim |------|   
x->oo\  x   /

\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}

Limit(((-2 + x)/x)^(2*x), x, oo, dir='-')

Detail solution

Let's take the limit

\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}

transform

\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}

\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x}

\lim_{x \to \infty} \left(- \frac{2}{x} + \frac{x}{x}\right)^{2 x}

\lim_{x \to \infty} \left(1 - \frac{2}{x}\right)^{2 x}

=
do replacement

u = \frac{x}{-2}

then

\lim_{x \to \infty} \left(1 - \frac{2}{x}\right)^{2 x}

=
=

\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}

\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{- 4 u}

\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-4}

The limit

\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}

is second remarkable limit, is equal to e ~ 2.718281828459045
then

\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{-4} = e^{-4}

The final answer:

\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x} = e^{-4}

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

Plot the graph

Rapid solution [src]

 -4
e

e^{-4}

Expand and simplify

Other limits x→0, -oo, +oo, 1

\lim_{x \to \infty} \left(\frac{x - 2}{x}\right)^{2 x} = e^{-4}

\lim_{x \to 0^-} \left(\frac{x - 2}{x}\right)^{2 x} = 1

\lim_{x \to 0^+} \left(\frac{x - 2}{x}\right)^{2 x} = 1

\lim_{x \to 1^-} \left(\frac{x - 2}{x}\right)^{2 x} = 1

\lim_{x \to 1^+} \left(\frac{x - 2}{x}\right)^{2 x} = 1

\lim_{x \to -\infty} \left(\frac{x - 2}{x}\right)^{2 x} = e^{-4}

The graph