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

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In the statement of the Lemma you start with $y_1$, but in the proof you refer to $y_0$. – Patrick Apr 6 '12 at 0:26
What happened to your previous question? Do you still need help with it? – David Mitra Apr 6 '12 at 13:40
In the future, please don't use the edit feature to change a question to a completely different one. Instead, ask a new question. You can then delete the previous question if you want (unless it already has answers). – Nate Eldredge Apr 6 '12 at 14:46

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 $$\lim \left[\frac{x}{x+2}\right]^{x/8} = \lim\left[\frac{1}{\frac{x+2}{x}}\right]^{x/8} =\frac{1}{\lim [1 + (2/x)]^x]^{1/8}}=\frac{1}{(e^2)^{1/8}}=e^{-1/4}.$$

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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 $\lim\limits_{x\rightarrow\infty}\ln({x\over x+2})^{x/8}=\lim\limits_{x\rightarrow\infty}\Bigl({x\over 8}\ln({x\over x+2}) \Bigr)$.

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