# Question #3f226

Jan 25, 2017

${\lim}_{x \to \infty} {\left(\frac{x}{2 + x}\right)}^{x - 2} = \frac{1}{e} ^ 2$

#### Explanation:

Write the expression as:

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

Now take the logarithm of this expression:

$\ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right) = \left(2 - x\right) \ln \left(1 + \frac{2}{x}\right) = 2 \ln \left(1 + \frac{2}{x}\right) - x \ln \left(1 + \frac{2}{x}\right)$

We can see that:

${\lim}_{x \to \infty} 2 \ln \left(1 + \frac{2}{x}\right) = 2 \ln \left(1\right) = 0$

so we can ignore this term.

Focusing on the other addendum we have:

$x \ln \left(1 + \frac{2}{x}\right) = 2 \ln \frac{1 + \frac{2}{x}}{\frac{2}{x}}$

Substituting $y = \frac{2}{x}$ we have ${\lim}_{x \to \infty} y \left(x\right) = 0$ so that:

${\lim}_{x \to \infty} \ln \frac{1 + \frac{2}{x}}{\frac{2}{x}} = {\lim}_{y \to 0} \ln \frac{1 + y}{y} = 1$

(you can find the explanation here )

Putting this together we can see that:

${\lim}_{x \to \infty} \ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right) = {\lim}_{x \to \infty} 2 \ln \left(1 + \frac{2}{x}\right) - x \ln \left(1 + \frac{2}{x}\right) = 0 - 2 \cdot 1 = - 2$

Now note that:

${e}^{\ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right)} = {\left(\frac{x}{2 + x}\right)}^{x - 2}$

so that:

${\lim}_{x \to \infty} {\left(\frac{x}{2 + x}\right)}^{x - 2} = {\lim}_{x \to \infty} \left({e}^{\ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right)}\right)$

and as ${e}^{x}$ is continuous in all of $\mathbb{R}$:

${\lim}_{x \to \infty} {\left(\frac{x}{2 + x}\right)}^{x - 2} = {e}^{{\lim}_{x \to \infty} \left(\ln \left({\left(\frac{x}{2 + x}\right)}^{x - 2}\right)\right)} = {e}^{- 2} = \frac{1}{e} ^ 2$

Jan 25, 2017

$\text{The Lim.=} \frac{1}{e} ^ 2.$

#### Explanation:

We will use the following Standard Form of Limit :

${\lim}_{t \to \infty} {\left(1 + \frac{1}{t}\right)}^{t} = e . \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(\star\right)$

$\text{Let, "x-2=y," so, "x=2+y." Also, as } x \to \infty , y \to \infty .$

$\text{The Reqd. Lim., now,=} {\lim}_{y \to \infty} {\left\{\frac{2 + y}{2 + \left(2 + y\right)}\right\}}^{y}$

$= {\lim}_{y \to \infty} {\left\{\frac{2 + y}{4 + y}\right\}}^{y}$

$= {\lim}_{y \to \infty} {\left\{\frac{y \left(\frac{2}{y} + 1\right)}{y \left(\frac{4}{y} + 1\right)}\right\}}^{y}$

${\lim}_{y \to \infty} {\left\{\frac{1 + \frac{2}{y}}{1 + \frac{4}{y}}\right\}}^{y}$

$= \frac{{\lim}_{y \to \infty} {\left(1 + \frac{2}{y}\right)}^{y}}{{\lim}_{y \to \infty} {\left(1 + \frac{4}{y}\right)}^{y}}$

$= {\left\{{\lim}_{y \to \infty} {\left(1 + \frac{2}{y}\right)}^{\frac{y}{2}}\right\}}^{2} / {\left\{{\lim}_{y \to \infty} {\left(1 + \frac{4}{y}\right)}^{\frac{y}{4}}\right\}}^{4}$

$= {e}^{2} / {e}^{4.} \ldots \ldots \ldots \ldots . \left[\because , \left(\star\right)\right]$

$\text{The Lim.=} \frac{1}{e} ^ 2 ,$ as Respected Andrea S. Sir has derived.