# Help solving the limit when $\lim_{x\to 1}{(\ln x)^{x-1}}$

I'm stuck with this one, so far I've tried taking logarithms to apply "The limit of the logarithm is the $ln$ of the limit" And then turning it into an infinite/infinite indeterminate but keep obtaining $0$ D:

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The limit of the log is $0$, so if $L$ is the desired limit, $\ln L=0$, and $L=~?$ – Brian M. Scott Jan 7 '13 at 22:49
Presumably you want the limit as $x$ approaches $1$ from the right, because of difficulties with the meaning of $a^b$ when $a$ is negative. – André Nicolas Jan 7 '13 at 22:57

## 3 Answers

Your limit is of the form $0^0$. In this case, it is helpful to use the fact that $a^b = \exp(b \ln a)$. You then have that (exp is continuous) $$\lim_{x\to 1} (\ln x )^{x-1} = \exp\left[ \lim_{x\to 1} (x-1) \ln\ln(x)\right] =\exp(0) =1$$ because $\ln x$ grows slower than any polynomial.

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L'Hospital's also a possibility:

$$(\log x)^{x-1}=e^{(x-1)\log\log x}$$

and

$$\lim_{x\to 1^+}(x-1)\log\log x=\lim_{x\to 1^+}\frac{\log\log x}{\frac{1}{x-1}}\stackrel{\text{L'H}}=\lim_{x\to 1^+}-\frac{(x-1)^2}{x\log x}\stackrel{\text{L'H}}=$$

$$=\lim_{x\to 1^+}-2\frac{x-1}{\log x+1}=-2\frac{0}{0+1}=0$$

Thus the limit is, by continuity of the exponential function, $\,e^0=1\,$

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Let us first set $u=x-1$. Note that your limit requires $x-1=u>0$ to make sense.

So your limit is equal to $$\lim_{u\rightarrow 0^+} (\ln (1+u))^u.$$ Now for all $u>0$, we have $$\ln(1+u)\leq u$$ since the graph of $\ln (1+u)$ is below the tangent at $u=0$ (by a concavity argument).

Using the fact that $\ln$ is increasing on $]0,+\infty[$, we have $$1\leq (\ln (1+u))^u=\exp\left( u\ln(\ln (1+u))\right)\leq \exp(u\ln u)$$ for all $u>0$.

It remains to invoke the Squeeze theorem and the fact that $$\lim_{u\rightarrow 0^+} u\ln u=0$$ to deduce from the inequality above that your limit is equal to $\exp(0)=1$.

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