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I came across an interesting limit I could not solve:

$$ \lim_{x \to 0^{+}}\left[\arcsin\left(x\right)\right]^{\tan\left(x\right)} $$

Given we have not proven l'Hôpital's rule yet, I have to solve it without it. Also, I would rather not use advanced methods such as the taylor series (which yield $x^x$ here).

Squeeze theorem does not (easily?) really help here, nor does the exponent function as far as I see it:

$$ \lim_{x\rightarrow 0+}(\arcsin x)^{\tan\,x} = \lim_{x\rightarrow 0+} e^{{\tan(x)}\ln(\arcsin x)} $$ Here again the exponent is an undefined term $(0 \cdot +\infty)$. Unlike all limits I practiced on however, this logarithm does not tend to $1$, so I don't really see how it cancels out.

Is there an easy solution I am missing?

Felix Marin
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Mikulas Dite
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2 Answers2

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$$\begin{align} \lim \limits_{x\to0^{+}}(\arcsin(x))^{\tan(x)}&=\lim \limits_{x\to0^{+}}e^{\large{\tan(x)\ln(\arcsin(x))}}\\ =e^{\lim_{x\to0^{+}}\frac{\tan(x)}{x}(\arcsin(x)\ln(\arcsin(x)))\frac{x}{\arcsin(x)} }=1\end{align}$$

Notice that $\lim_{x\to0^{+}}\arcsin(x)\ln(\arcsin(x))=-\lim_{x\to0^{+}}\frac{\ln(\frac{1}{\arcsin(x)})}{{\frac{1}{\arcsin(x)}}}=-\lim_{u\to\infty}\frac{\ln(u)}{u}$

We demonstrate this limit by noticing: $0\le\frac{\ln(u)}{u}=\frac{1}{u}\int_{1}^{u}\frac{1}{t}dt\le\frac{1}{u}\int_{1}^{u}\frac{1}{\sqrt{t}}dt=\frac{2(\sqrt{u}-1)}{u}=2(\frac{1}{\sqrt{u}}-\frac{1}{u})$

for $u$ sufficiently large. So the limit is $0$.

$\lim_{x\to0^{+}}\frac{x}{\arcsin(x)}=\frac{1}{\frac{d(\arcsin(x))}{dx}}|_{x=0^{+}}=\sqrt{1-x^{2}}|_{x=0^{+}}=1$.

Also $\lim_{x\to0^{+}}\frac{\tan(x)}{x}=\lim_{x\to0^{+}}\frac{\sin(x)}{x}\frac{1}{\cos(x)}=1$.

user71352
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  • Why would the log cancel out? As far as I remember $\lim\limits_{x\rightarrow 0}\frac{\log(x+1)}{x} = 1$, but this does not seem to be the case. – Mikulas Dite Dec 29 '13 at 21:08
  • Sorry I was in the middle of making corrections. – user71352 Dec 29 '13 at 21:11
  • Bear with me for a second here please: $(1 + \arcsin(x) - 1) \rightarrow 0 \ne 1$, so why did the log cancel out? Are you using the limit I wrote in the comment above? Because $\lim_{x\to 0} , \frac{\log \left(\arcsin(x)\right)}{\arcsin(x)}=-\infty$ – Mikulas Dite Dec 29 '13 at 21:22
  • I am using the limit that you wrote in your comment where $x$ will be denoted by $\arcsin(x)-1$. – user71352 Dec 29 '13 at 21:26
  • However, if $s = \arcsin(x)-1 \wedge x\rightarrow 0 \Rightarrow s \rightarrow -1$, because $\arcsin(0)=0$, right? Meaning the whole argument of the log tends to $0$, not $1$, thus the partial limit in the topmost comment above does not seem applicable. – Mikulas Dite Dec 29 '13 at 21:28
  • I just came back here for this very reason... It seems your argument is wrong. :-( – Lucian Dec 29 '13 at 21:36
  • I have been made aware I'm just thinking of how to fix it right now. My apologies. – user71352 Dec 29 '13 at 21:37
  • Seems valid now. In the last part I will rather use $\lim_{+\infty} \frac{\ln u}{u} = \ln \lim_{+\infty} n^{1/n} = \ln 1 = 0$ rather than the integral however, taking the last limit axiomatic as I have proven that before. Thanks :) – Mikulas Dite Dec 29 '13 at 22:01
  • Your welcome. Sorry for the earlier mistake. If you catch any more I am more than happy to correct it. – user71352 Dec 29 '13 at 22:08
  • ...or just replace u with $e^t$... – Lucian Dec 29 '13 at 22:10
  • I'm sorry, I just had to pick the other answer as it's more elegant. I do really appreciate the effort though, sir. I have learned a lot through our discussion. – Mikulas Dite Dec 29 '13 at 22:25
  • No problem happy to help. Glad you have learned through the discussion. – user71352 Dec 29 '13 at 22:28
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$$\lim \limits_{x\to0^{+}}(\arcsin(x))^{\tan(x)}=\lim \limits_{x\to0^{+}}e^{\large{\tan(x)\ln(\arcsin(x))}}$$ therefore, $\tan(x)\ln(\arcsin(x))=\tan(x)\ln\left[x(1+\frac{\arcsin(x)}{x}-1)\right]=$ $$\tan(x)\ln(x)+\tan(x)\frac{\ln(1+\frac{\arcsin(x)}{x}-1)}{\frac{\arcsin(x)}{x}-1} \left (\frac{\arcsin(x)}{x} -1\right)$$ the second addend goes to zero as $x \rightarrow 0^{+}$, because $(\frac{\arcsin(x)}{x}-1)$ and $\tan(x)$ $\rightarrow 0^{+}$ as $x \rightarrow 0^{+}$ while the central factor tends to 1. The first addend is equal to $x\ln(x)\frac{\tan(x)}{x}$ which tends to zero. The whole exponent tends to zero, therefore $$\lim \limits_{x\to0^{+}}e^{\large{\tan(x)\ln(\arcsin(x))}}=e^0=1$$

Lucian
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Matheman
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