Mister Exam

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

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

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

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

Detail solution

Let's take the limit

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

transform

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

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

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

\lim_{x \to \infty} \left(1 + \frac{4}{x - 2}\right)^{x}

=
do replacement

u = \frac{x - 2}{4}

then

\lim_{x \to \infty} \left(1 + \frac{4}{x - 2}\right)^{x}

=
=

\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{4 u + 2}

\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{2} \left(1 + \frac{1}{u}\right)^{4 u}\right)

\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{2} \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 - 2}\right)^{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 - 2}\right)^{x} = e^{4}

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

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

\lim_{x \to 1^-} \left(\frac{x + 2}{x - 2}\right)^{x} = -3

\lim_{x \to 1^+} \left(\frac{x + 2}{x - 2}\right)^{x} = -3

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

The graph