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How can I find the limit : $\lim\limits_{x\to\infty}(1+x)^{\frac{1}{x}}$

I know I can use the following statement:

$$(1+x)^{1/x}=e^{\frac 1 x\log(1+x)}$$

but I don't know how to continue from there

(I'm still not allowed to use Lhoptial if that could be used here)

2 Answers 2

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Observe \begin{align} \lim_{x\rightarrow \infty} \exp\left( \frac{\log(1+x)}{x}\right)= \exp\left( \lim_{x\rightarrow \infty}\frac{\log(1+x)}{x}\right). \end{align} Since \begin{align} \lim_{x\rightarrow \infty}\frac{\log(1+x)}{x} = 0 \end{align} then it follows \begin{align} \lim_{x\rightarrow \infty} \exp\left( \frac{\log(1+x)}{x}\right)=1 \end{align}

Note \begin{align} \log(1+x) \leq \sqrt{x} \end{align} when $x>0$.

Additional:

The definition of $\log(1+x)$ is defined to be \begin{align} \log(1+x) := \int^x_0 \frac{dt}{1+t} \end{align} and since \begin{align} \sqrt{x} = \int^x_0 \frac{dt}{\sqrt{t}} \end{align} then we have \begin{align} \sqrt{x}-\log(1+x) = \int^x_0 \frac{1}{\sqrt{t}}-\frac{1}{t+1}\ dt\geq 0 \end{align} since \begin{align} 1+t\geq \sqrt{t} \ \ \Leftrightarrow 1-\sqrt{t}+t = \frac{3}{4}+\left(\sqrt{t}-\frac{1}{2}\right)^2\geq 0. \end{align}

  • 0
    How would you prove that $\lim_{x\to +\infty} \frac{\log(1+x)}{x}=0$? – 2017-01-09
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    @flytothesurface I hope my post have answered your question. – 2017-01-09
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    Thank you. Actually it does, but I still don't know a way of proving that inequality without involving derivatives – 2017-01-09
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$$A=\lim_{x \to \infty} (1+x)^{\frac{1}{x}}\\ \ln A=\ln \lim_{x \to \infty} (1+x)^{\frac{1}{x}}\\= \lim_{x \to \infty} \ln (1+x)^{\frac{1}{x}}=\\ \lim_{x \to \infty} \frac{1}{x}\ln(1+x)=\\ \lim_{x \to \infty} \frac{\ln(1+x)}{x}=\\ \lim_{x \to \infty} \frac{\frac{1}{1+x}}{1}=0\\ \ln A=0 \to A=e^0=1 $$

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    Could you explain why : $\lim_{x \to \infty} \frac{\ln(1+x)}{x}= \lim_{x \to \infty} \frac{\frac{1}{1+x}}{1}$ – 2017-01-09
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    I used hop rule , and $$(\ln(f(x))'=\frac{f'}{f} \to (\ln(1+x))'=\frac{1}{1+x}$$ – 2017-01-09