Mathematics - Limits Question with Solution | TestHub

$$\underset{x\to \infty }{\lim }{\left(\frac{|x|}{|x|+2}\right)}^{-x}$$

$$=\underset{x\to -\infty }{\lim }{\left(\frac{2-x-2}{2-x}\right)}^x$$

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

As $x\to -\infty$, $|x|=-x$.

Let $x=-\frac{1}{y}$, so $y\to 0$.

$${\mathrm{Lt}}_{y\to 0}{\left(1-\frac{2}{2+\frac{1}{y}}\right)}^{-\frac{1}{y}}$$

$$={\mathrm{Lt}}_{y\to 0}{\left(1-\frac{2y}{2y+1}\right)}^{-\frac{1}{y}}$$

This is of the $1^{\infty }$ form.

$$=e^{{\mathrm{Lt}}_{y\to 0}\left(-\frac{1}{y}\right)\left(1-\frac{2y}{2y+1}-1\right)}$$

$$=e^{{\mathrm{Lt}}_{y\to 0}\left(-\frac{1}{y}\right)\left(-\frac{2y}{2y+1}\right)}$$

$$=e^{{\mathrm{Lt}}_{y\to 0}\left(\frac{2}{2y+1}\right)}$$

$$=e^{\frac{2}{2(0)+1}}=e^2$$