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I want to find $\lim \limits_{x\to \infty}(e^{2x}+1)^{1/x}$.

The first thing I thought is that $\lim \limits_{x\to\infty}\frac{1}{x}=0$.

So the limit would be $\lim \limits_{x\to \infty}(e^{2x}+1)^{1/x}= 1$ but I am pretty sure that this is not the right answer.

Then I read in wikipedia that I can apply L'Hospital on $\infty^{0}$.

The only problem is that I don't know how to do it.

Do I have to transform it in something like $\infty^{\infty}$ or $0^{0}$?

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5 Answers 5

up vote 7 down vote accepted

We have $e^{2x} < e^{2x} + 1 < e^{2x} + e^{2x}$ for $x > 0$. Hence, we get $$\left(e^{2x} \right)^{1/x} < \left(e^{2x} +1 \right)^{1/x} < \left(e^{2x} + e^{2x} \right)^{1/x}$$ i.e. $$e^2 < \left(e^{2x} +1 \right)^{1/x} < \left(2e^{2x} \right)^{1/x} = 2^{1/x} e^2$$ Now take the limit and note that $\displaystyle \lim_{x \rightarrow \infty} 2^{1/x} = 1$ to get that $$e^2 \leq \lim_{x \rightarrow \infty} \left(e^{2x} +1 \right)^{1/x} \leq e^2$$ Hence, we get that $$\lim_{x \rightarrow \infty} \left(e^{2x} +1 \right)^{1/x} = e^{2}$$

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Isn't $\lim_{x \rightarrow \infty} 2^{1/x} = 1$? –  pressy_paris Dec 10 '11 at 12:14
    
aah. I am sorry. It was a typo. The rest of the argument stays fine as it is. –  user17762 Dec 10 '11 at 12:15
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We have, after taking the logarithm, an indeterminate form $0\cdot \infty$ (which can be different from $0$, for example taking $f(x)=1/x$ and $g(x)=ax$ where $a>0$). Write $\left(e^{2x}+1\right)^{\frac 1x}=e^2\left(1+e^{-2x}\right)^{\frac 1x}=e^2\exp\left(\frac 1x\ln (1+e^{-2x})\right)$, and we don't have anymore an indeterminate form since $\lim_{x\to +\infty}\frac 1x\ln (1+e^{-2x})=0$.

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Very nice, no L'Hopital :) –  David Mitra Dec 10 '11 at 12:37
    
$\lim\limits_{x\to\infty}\left(1+e^{-2x}\right)^{\frac{1}{x}}=1^0$, so you could stop there. :-) –  robjohn Dec 12 '11 at 1:15
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[Edit] The other answers are much better. But, as this falls into a standard class of problems in the application of L'Hopital's rule, I'll offer:

If you have an $\infty^0$ form, it is indeterminate (for example $\lim\limits_{x\rightarrow\infty}(e^x)^{1/x}=e$, but $\lim\limits_{x\rightarrow\infty}((e^x)^x)^{1\over x}=\infty$).

To handle a form of this type, you may take the limit of the logarithm of the expressionn first. Then, using the power rule for logarithms, you can set things up so that L'Hopital's rule is applicable. After you find the limit of the logarithm, say it's $L$, then the limit of the original expression will be $e^L$.

For your example:

Find the limit of the logarithm first.

Set $$g(x)=\ln\bigl( (e^{2x}+1)^{1/x}\bigr)= \underbrace{{1\over x}\ln( e^{2x}+1)}_{0\cdot\infty\text{ form}}= \underbrace{{\ln (e^{2x}+1)\over x}}_{{\infty\over\infty}\text{ form}}.$$

Then, using L'Hopital: $$\lim_{x\rightarrow\infty} g(x)= \lim_{x\rightarrow\infty}{\ln(e^{2x}+1)\over x}=\lim_{x\rightarrow\infty}{2e^{2x} \over e^{2x}+1} =\lim_{x\rightarrow\infty}{4e^{2x} \over 2e^{2x} } =2. $$

So $$\eqalign{ \lim_{x\rightarrow\infty}{(e^{2x}+1 )^{1/x}} &=\lim_{x\rightarrow\infty}{ \exp\bigl[\, g(x) \bigr]}\cr &={ \exp\bigl[\, \lim_{x\rightarrow\infty} g(x) \bigr]}\cr & =e^2.} $$

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Noting that $(1 + e^{2x})^\frac{1}{x} = (1 + e^{2x})^\frac{2}{2x}$, one can define $t = 2x$ and see that the limit is the square of $e (1 + e^{-t})^\frac{1}{t}$, in which $1 + e^{-t}$ clearly tends to 1+ as t tends to infinity, and hence a fortiori so does $(1 + e^{-t})^\frac{1}{t}$.

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In logarithm formulas we have: $$x = e^{\ln{x}}$$

we use this formula for converting $\infty^{0}$ to $0 \times \infty$.

so we go on:

$\lim \limits_{x\to \infty}(e^{2x}+1)^{1/x} = \lim \limits_{x\to \infty} e^{\ln{(e^{2x}+1)^{1/x}}} = \lim \limits_{x\to \infty} e^{\frac{1}{x} \times \ln{(e^{2x}+1)}} = \lim \limits_{x\to \infty} e^{\frac{\ln{(e^{2x}+1)}}{x}} $

now it became $\frac{\infty}{\infty}$ and you can apply L'Hospital's rule.

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