How to transform this limit expression as a limit of $e$ I have the following expression:
$\displaystyle\lim _{x\to \infty }\left(\dfrac{x+2}{\:x-6}\right)^\left(\dfrac{x}{\:4}\right)$
I'm studying Calculus I and our lector has shown us ways of transforming such limits to:
$\displaystyle\lim_{x\to \infty }\left(1+\frac{1}{\:x}\right)^x = e$
The way this calculator solves it is not immediately clear to me, is there any other way to find the above limit?
 A: HINT
We have that
$$\left(\frac{x+2}{x-6}\right)^{\frac x 4}=\left(\frac{x-6+8}{x-6}\right)^{\frac x 4}=\left(1+\frac{8}{x-6}\right)^{\frac x 4}$$
then we can manipulate further in order to use the standard limit.
Refer to the related


*

*Calculating a limit with trignonometeric and quadratic function
A: You may proceed as follows:


*

*Set $y = x-6$
$$\left(\frac{x+2}{x-6} \right)^{\frac{x}{4}} = \left(1 +\frac{8}{y} \right)^{\frac{y+6}{4}} = \left(1 +\frac{2}{\frac{y}{4}} \right)^{\frac{y}{4}}\cdot \left(1 +\frac{8}{y} \right)^{\frac{3}{2}} \stackrel{y \to \infty}{\longrightarrow}e^2$$
A: $\displaystyle\lim _{x\to \infty }\left(\dfrac{x+2}{\:x-6}\right)^\left(\dfrac{x}{\:4}\right)=\lim _{x\to \infty }\left(\dfrac{1+\dfrac2x}{1-\dfrac6x}\right)^\left(\dfrac{x}{\:4}\right)=\dfrac{\left(\lim_{x\to\infty}\left(1+\dfrac2x\right)^{x/2}\right)^{1/2}}{\left(\lim_{x\to\infty}\left(1-\dfrac6x\right)^{-x/6}\right)^{-3/2}}=\dfrac{e^{1/2}}{e^{-3/2}}=?$
A: If you divide $x+2$ by $x-6$ the quotient is $1$ and the remainder is $8$, so the limit becomes
$$\lim_{x\to\infty}\left(1+\frac{8}{x-6}\right)^{x/4}$$
Now, substitute $x/4$ by
$$\frac{x-6}8\cdot\frac{8}{x-6}\cdot\frac x4$$
to get
$$\lim_{x\to\infty}\left[\left(1+\frac{1}{\dfrac{x-6}8}\right)^{\dfrac{x-6}8}\right]^{\dfrac{8x}{4(x-6)}}$$
