# Evaluating $\lim_{x \to 0^+} (e^x-1)^{\frac{(\tan{x})^2}{\sqrt[3]{x^2}}}$

I cannot figure this limit out.

$$\lim_{x \to 0^+} (e^x-1)^{\frac{(\tan{x})^2}{\sqrt[3]{x^2}}}$$

I've used the e to the ln trick and multiplied by 1 ($\frac{x^2}{x^2}$) and arrived at

$$\lim_{x \to 0^+} \exp({x^{4/3}} \ln ({e^x-1}))$$

However I failed at getting further. I tried adding and subtracting $\ln x$ but that got me nowhere.

I cannot use l'Hospital or Taylor series (only the "known" limits for $\sin$, $\cos$, $e^x$, $\ln$ such as $\lim_{x \to 0}\frac{sinx}{x}=1$ which are really only Taylor series).

Thanks for help!

• Apart from well known limits $\lim_{x \to 0}\dfrac{\sin x}{x} = 1, \lim_{x \to 0}\dfrac{e^{x} - 1}{x} = 1, \lim_{x \to 0}\dfrac{\log (1 + x)}{x} = 1$ you will also need the lesser well known limit $\lim_{x \to \infty}\dfrac{\log x}{x^{a}} = 0$ if $a > 0$. Dec 6 '13 at 4:03

Hint: $$\lim_{x\to 0}\frac{e^x-1}x=\lim_{x\to 0}\frac{\tan x}x=1$$

• That's what I meant by adding and subtracting the logarithm in there - I would end up with this limit inside the logarithm and then could even get rid of the logarithm alltogether. I would still end up with a $\ln x$ there which doesn't seem to help. edit: or perhaps I should take the hint differently?
– Dahn
Dec 5 '13 at 19:00
• surely you can evaluate $\lim_{x\to0}x^{4/3}\log x$ Dec 5 '13 at 19:06
• I suppose that is why I got so stuck at such an elementary problem as this. No, I cannot, I feel ashamed of myself! What am I missing?
– Dahn
Dec 5 '13 at 19:24
• Sorry if that came off as arrogant (or something similar). Usually you prove it by substituting $u=\frac1x$, which obtains $$\lim_{u\to\infty} \frac{-\log u}{u^{4/3}}$$ It should be pretty clear that that evaluates to $0$ Dec 5 '13 at 19:28
• Oh of course, the limit goes to zero from right. Thanks! edit: I decided to acccept this as the answer mostly because of the quick response in comments. Thanks again.
– Dahn
Dec 5 '13 at 19:31

Replace tan x by sinx/cosx, ln(e^x-1) by ln(x) and add the sin x/x somewhere and you should see the answer. Only difficulty is in showing the equivalence of ln(e^x-1) and ln(x) but you can show that 1+x < e^x < 1+x+x^2 by hand if x is small enough no ?

• If I do that, won't I just end up with $\lim_{x \to 0^+} \exp ({x^{\frac{4}{3}}\ln x})$?
– Dahn
Dec 5 '13 at 19:02
• exactly. and x ln x -> 0, it is a common limit to know Dec 5 '13 at 21:46
• Just an additional question: does it exist if x goes to zero (i.e. not just from right)?
– Dahn
Dec 5 '13 at 23:28
• ln(x) is not defined in x=0. All you can do is define a function f(x) = 0 if x=0 and f(x)=x ln(x) if x >0. f is continuous Dec 6 '13 at 9:18

Hint:

$$L=\lim_{x\to 0^+} (e^x-1)^{\frac{(\tan x)^2}{\sqrt[3]{x^2}}}=\exp\left\{\lim_{x\to 0^+}\frac{\tan^2x}{\sqrt[3]{x^2}}\ln(e^x-1)\right\}=\exp\left\{\lim_{x\to 0^+}\frac{\tan^2x}{x^2} \sqrt[3]{\left(\frac{x}{e^x-1}\right)^4} \left((e^x-1)^{\frac{4}{3}}\ln(e^x-1)\right)\right\}=\exp\left\{0\right\}=1$$

My result: $L=1$

Compute the limit of the logarithm instead. This can be written as $\lim_{x\to0^+}\left(\dfrac{\tan x}{x}\right)^2x^{4/3}\log(e^x-1)$. The squared factor tends to 1 and the remaining product tends to $-\infty$. So the limit of the log is $-\infty$. Conclude that the original limit is 0.