2
$\begingroup$

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

$\ln{y}=x\ln{(\ln{(1+x)})}$

$\ln{y}=x\ln{(1+\ln{(1+x)}-1)}$

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

$\ln{y}=O{(x)}$

with this then

$e^{ln{y}}=e^{O{(x)}}$

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

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

$\endgroup$
  • 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
0
$\begingroup$

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 $\lim_{x \to 0^{+}} y = e^{L} = 1$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.