Evaluate the following limit:

$\lim\limits_{x \to 0^{+}} [\ln{(1+x)}]^{x}$

This is what I have, and I get the "correct" answer- just want to make sure my reasoning is valid:

let $y=[\ln{(1+x)}]^{x}$



Taylor expand $\ln{(1+x)}$ (the outermost one, where $x=(1+\ln{(1+x})-1)$

$\ln{y}=x(\ln{(1+x)}-1)-\frac{(\ln{(1+x)}-1)^{2}}{2} + H.O.T.$

$\ln{y}=x((x-\frac{x^{2}}{2}+H.O.T.)-1) - \frac{((x-\frac{x^{2}}{2} + H.O.T.)-1)^{2}}{2} + H.O.T.$

At this point, hopefully it is clear that everything will be multiplied out leaving


with this then


$\lim\limits_{x \to 0^{+}} y=e^{O{(x)}}=1$

Anyone see a mistake I'm making or want to poke a hole in this method?

  • 3
    $\begingroup$ You could just apply L'Hospital rule after taking logarithm of both sides. This is much easier. $\endgroup$ – ThePortakal Jun 13 '14 at 0:56

You have taken the right approach and the only issue is to calculate $$L = \lim_{x \to 0^{+}}x\log(\log(1 + x))$$ which can be done very easily as follows $$\begin{aligned}L &= \lim_{x \to 0^{+}}x\log(\log(1 + x))\\ &= \lim_{x \to 0^{+}}x\log(\log(1 + x))\\ &= \lim_{x \to 0^{+}}x\log\left(x\cdot\frac{\log(1 + x)}{x}\right)\\ &= \lim_{x \to 0^{+}}x\log x + x\log \left(\frac{\log(1 + x)}{x}\right)\\ &= \lim_{x \to 0^{+}}x\log x + 0\cdot\log 1\\ &= \lim_{x \to 0^{+}}x\log x\\ &= \lim_{t \to \infty}-\frac{\log t}{t}\text{ where } t = 1/x\\ &= 0\end{aligned}$$ Here we have used two standard logarithmic limits namely $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1\text{ and }\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0\text{ for any }a > 0$$ Now as you have done $\lim_{x \to 0^{+}} y = e^{L} = 1$.

  • $\begingroup$ Ah. I'm not familiar with those standard log limits. I'll be sure to remember those going forward. Thanks! $\endgroup$ – Adam Jun 14 '14 at 11:27

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