Limit and continuity For this question, should I use differentiation method or the integration method ?
$\lim_{x\to \infty} (\frac{x}{x+2})^{x/8}$
this is what i got so far: 
Note: $\lim \limits_{n\to\infty} [1 + (a/n)]^n = e^{\underline{a}}\ldots\ldots (1)$
$$
L = \lim \left[\frac{x}{x+2}\right]^{x/8} = \lim\left[\frac{1}{\frac{x+2}{x}}\right]^{x/8} =\frac{1}{\lim [1 + (2/x)]^x]^{1/8}}
$$

but i'm not sure where to go from there 
 A: I'm not sure what you mean by "differentiation or integration method" but:
Do you know that your note can be slightly generalized to $\lim\limits_{x\rightarrow\infty} (1+{a\over x})^{ x} =e^a$? If so, then just write $\lim\limits_{x\rightarrow\infty} \bigl((1+{2\over x})^{ x} \bigr)^{1/8}=\Bigl(\lim\limits_{x\rightarrow\infty}  (1+{2\over x})^{ x} \Bigr)^{1/8} $.
If not you could use L'Hôpital's rule to first evaluate $\lim\limits_{x\rightarrow\infty}\ln({x\over x+2})^{x/8}=\lim\limits_{x\rightarrow\infty}\Bigl({x\over 8}\ln({x\over x+2}) \Bigr) $.
A: You are probably intended to use the fact that
$$\lim_{t\to\infty}\left(1+\frac{1}{t}\right)^t=e.$$
A manipulation close to what you were doing gets us there. 
We have 
$$\left(1+\frac{2}{x}\right)^{x/8}=\left(\left(1+\frac{2}{x}\right)^{x/2}\right)^{1/4}.$$
Let $t=\frac{x}{2}$. Then $\frac{2}{x}=\frac{1}{t}$. Note that as $x\to\infty$, we have $t\to\infty$. It follows that
$$\lim_{x\to \infty}\left(1+\frac{2}{x}\right)^{x/8}=\lim_{t\to\infty}\left(\left(1+\frac{1}{t}\right)^{t}\right)^{1/4}=e^{1/4}.$$ 
So our limit is $1/e^{1/4}$, or equivalently $e^{-1/4}$.
If you are allowed to use the fact that in general $\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x=e^a$, then you can simply take $a=2$, and conclude that
$$\lim \left[\frac{x}{x+2}\right]^{x/8} = \lim\left[\frac{1}{\frac{x+2}{x}}\right]^{x/8} =\frac{1}{\lim [1 + (2/x)]^x]^{1/8}}=\frac{1}{(e^2)^{1/8}}=e^{-1/4}.$$
