Evaluate $\lim_{x \rightarrow 1^+} (\ln\ x)^{x-1}$

Question

Evaluate:

$$\lim_{x \rightarrow 1^+} (\ln\ x)^{x-1}$$

Answer 1

HINT:

$$\frac{x-1}{x}\le\log(x)\le x-1$$

and

$$\lim_{x\to 0^+}x^x=1$$

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SPOILER ALERT: Scroll over the highlighted area to reveal the solution

In THIS ANSWER and THIS ONE, I showed using only (i) the limit definition of the exponential function and (ii) Bernoulli's Inequality that the logarithm function satisfies the inequalities $$\frac{x-1}{x}\le\log(x)\le x-1$$for $x>0$. Therefore, we can write $$\left(\frac{x-1}{x}\right)^{x-1}\le\left(\log(x)\right)^{x-1}\le (x-1)^{x-1} \tag 1$$Using the well-known result $\lim_{x\to 0^+}x^x=1$ we see that $\lim_{x\to 1^+}(x-1)^{x-1}=\lim_{x\to 1^+}\left(\frac{x-1}{x}\right)^{x-1}=1$. Applying the squeeze theorem to $(1)$ yields $$\lim_{x\to 1^+}\left(\log(x)\right)^{x-1}=1$$

Answer 2

HINT

Take logs to get $(x-1)\ln \ln x$ and use L'Hospital's Rule after a minor transformation.