Limit of $\lim_{x\to 0^+} (1+x)^{\ln x}$

Question

This is what i did:
$\lim_{x\to 0^+} (1+x)^{\ln x}=\lim_{x\to \infty}(1+\frac1x)^{\ln \frac1x}$
Then
$(1+\frac1x)^{\ln \frac1x}=(1+\frac1x)^{-\ln x}=(1+\frac1x)^{\frac xx(-\ln x)}=(1+\frac1x)^{\frac xx(-\ln x)}=[(1+\frac1x)^x]^{-\frac{\ln x}{x}}$
Then
$\lim_{x\to \infty}[(1+\frac1x)^x]^{-x\ln x}=\lim_{x \to \infty}e^{-\frac{\ln x}{x}}$
Then
$\lim_{x \to \infty}{-\frac{\ln x}{x}}=L'Hospital=\lim_{x\to\infty}\frac{-1}{x}=0$
Then
$\lim_{x \to \infty}e^{-\frac{\ln x}{x}}=e^0=1$

Is this process ok? Is there an easier way of doing this limit?

score 2 · Accepted Answer · answered Nov 19 '15 at 23:35

2

That is very complicated. You can have something much simpler using only basic limits and equivalents. First note it is enough to find the limit of $\ln(1+x)^{\ln x}$: as $\;\ln(1+x)\sim_0 x$, we have: $$\ln\bigl((1+x)^{\ln x}\bigr)=\ln x\ln(1+x)\sim_0x\ln x\xrightarrow[x\to 0+]{}0,$$ hence $\;\lim\limits_{x\to 0+}\ln(1+x)^{\ln x}=\mathrm e^0=1$.

answered Nov 19 '15 at 23:35

Bernard

175,478

Sorry, i cant understand how $\lim_{x\to 0^+} (1+x)^{\ln x}=\lim_{x\to0^+}\ln (1+x)^{\ln x}=e^0=1$ – Valentin Nov 19 '15 at 23:48
$\mathrm e^x$ is a continuous function. – Bernard Nov 19 '15 at 23:52
$\lim_{x\rightarrow 0^+} (1+x)^{\ln(x)}$ is not equal to $\lim_{x\rightarrow 0^+}\ln\left((1+x)^{\ln(x)}\right)$, it's equal to $e^x$ of that. That's what Bernard's solution is using. – Zavosh Nov 19 '15 at 23:57
I get it now, thanks a lot for your time both of you – Valentin Nov 20 '15 at 00:21
You're welcome! – Bernard Nov 20 '15 at 00:44

Limit of $\lim_{x\to 0^+} (1+x)^{\ln x}$

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