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?

  • $\begingroup$ When $x$ is close to $0$, both $\arcsin x$ and $\tan x$ are 'equal' to $x$, so you get $x^x$. (This is essentially the first order Taylor approximation.) $\endgroup$ – Ragnar Dec 29 '13 at 21:05
  • $\begingroup$ The limit should be $1$. $\endgroup$ – Mhenni Benghorbal Dec 29 '13 at 21:07
  • $\begingroup$ @Ragnar fair enough to be an intuition, I think am expected to give a more elaborate proof however. $\endgroup$ – Mikulas Dite Dec 29 '13 at 21:42
  • 1
    $\begingroup$ $\large\arcsin\left(x\right)$ and $\large\tan\left(x\right)$ $\large\sim x$ when $\large x \sim 0$. $\endgroup$ – Felix Marin Dec 29 '13 at 21:47

$$\lim \limits_{x\to0^{+}}(\arcsin(x))^{\tan(x)}=\lim \limits_{x\to0^{+}}e^{\large{\tan(x)\ln(\arcsin(x))}}$$ 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 $$\lim \limits_{x\to0^{+}}e^{\large{\tan(x)\ln(\arcsin(x))}}=e^0=1$$

  • $\begingroup$ Oh this is so awesome. Thank you. Love it! $\endgroup$ – Mikulas Dite Dec 29 '13 at 22:24
  • $\begingroup$ you're welcome! $\endgroup$ – Matheman Dec 29 '13 at 22:26

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

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 $0$.


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

  • $\begingroup$ 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. $\endgroup$ – Mikulas Dite Dec 29 '13 at 21:08
  • $\begingroup$ Sorry I was in the middle of making corrections. $\endgroup$ – user71352 Dec 29 '13 at 21:11
  • $\begingroup$ 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 $\lim_{x\to 0} \, \frac{\log \left(\arcsin(x)\right)}{\arcsin(x)}=-\infty$ $\endgroup$ – Mikulas Dite Dec 29 '13 at 21:22
  • $\begingroup$ I am using the limit that you wrote in your comment where $x$ will be denoted by $\arcsin(x)-1$. $\endgroup$ – user71352 Dec 29 '13 at 21:26
  • $\begingroup$ 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. $\endgroup$ – Mikulas Dite Dec 29 '13 at 21:28

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