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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!

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  • $\begingroup$ 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 $a > 0$. $\endgroup$ Dec 6 '13 at 4:03
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Hint: $$\lim_{x\to 0}\frac{e^x-1}x=\lim_{x\to 0}\frac{\tan x}x=1$$

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  • $\begingroup$ 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? $\endgroup$
    – Dahn
    Dec 5 '13 at 19:00
  • $\begingroup$ surely you can evaluate $\lim_{x\to0}x^{4/3}\log x$ $\endgroup$ Dec 5 '13 at 19:06
  • $\begingroup$ 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? $\endgroup$
    – Dahn
    Dec 5 '13 at 19:24
  • $\begingroup$ 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 $0$ $\endgroup$ Dec 5 '13 at 19:28
  • $\begingroup$ 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. $\endgroup$
    – Dahn
    Dec 5 '13 at 19:31
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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 ?

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  • $\begingroup$ If I do that, won't I just end up with $\lim_{x \to 0^+} \exp ({x^{\frac{4}{3}}\ln x})$? $\endgroup$
    – Dahn
    Dec 5 '13 at 19:02
  • $\begingroup$ exactly. and x ln x -> 0, it is a common limit to know $\endgroup$
    – Thomas
    Dec 5 '13 at 21:46
  • $\begingroup$ Just an additional question: does it exist if x goes to zero (i.e. not just from right)? $\endgroup$
    – Dahn
    Dec 5 '13 at 23:28
  • $\begingroup$ 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 $\endgroup$
    – Thomas
    Dec 6 '13 at 9:18
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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: $L=1$

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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.

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