I'm trying to solve this problem, $\lim_{x\to 0+}(\sin^2(4x))^{\sin^{-1}(2x)}$. It's indeterminate form so I used L'Hospital's rule but I'm stuck at here;

$\lim_{x\to 0+}\ln(\sin^2(4x))^{\sin^{-1}(2x)}=\lim_{x\to 0+} \frac{2\ln(\sin(4x))}{1/\sin^{-1}(2x)}=\lim_{x\to 0+}\frac{4(\sin^{-1}(2x))^2\sqrt{1-4x^2}}{\tan x} $

Could you help me? Thank you in advance.

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    $\begingroup$ I'm a bit lost on the work you've done already. The classical way to use l'Hospital in forms with an exponent is to calculate the limit of the natural logarithm of the form, which changes the exponent into a factor. $\endgroup$ – Alexander Geldhof Nov 11 '19 at 10:22
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    $\begingroup$ Does $\sin^{-1}(2x)$ mean $\arcsin(2x)$? $\endgroup$ – José Carlos Santos Nov 11 '19 at 10:23
  • $\begingroup$ @Alexander Geldhof Sorry for any confuse, I skipped some calculations. I used the fact that f(x)^g(x)=e^f(x)*g(x) and a*b=b*(1/a) and used l'Hospital in second equality. $\endgroup$ – Scott Lee Nov 11 '19 at 11:42
  • $\begingroup$ @José Carlos Santos Yes it does. $\endgroup$ – Scott Lee Nov 11 '19 at 11:42

Just use that $\lim_{t\to 0^+}t\ln t = 0$ as follows:

You may write

$$\ln\left((\sin^2(4x))^{\sin^{-1}(2x)}\right) =\underbrace{\frac{\sin^{-1} (2x)}{2x}}_{\stackrel{x\to 0^+}{\rightarrow}1}\cdot \underbrace{\frac{4x}{\sin 4x}}_{\stackrel{x\to 0^+}{\rightarrow}1}\underbrace{\sin (4x)\ln (\sin 4x)}_{\stackrel{x\to 0^+}{\rightarrow}0}$$

It follows $$\lim_{x\to 0+}(\sin^2(4x))^{\sin^{-1}(2x)}= e^0 = 1$$

  • $\begingroup$ The best way of course! $\endgroup$ – user Nov 11 '19 at 11:38
  • $\begingroup$ Thanks very much. I didn't know that limit arcsinx/x =1. $\endgroup$ – Scott Lee Nov 11 '19 at 11:44
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    $\begingroup$ @ScottLee : You are welcome. You can see this limit just by substituting $x= \sin t \Rightarrow \frac{t}{\sin t}$. $\endgroup$ – trancelocation Nov 11 '19 at 11:46
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    $\begingroup$ @user : Thanks user (gimusi) :-D $\endgroup$ – trancelocation Nov 11 '19 at 11:47
  • $\begingroup$ @trancelocation Awesome! $\endgroup$ – Scott Lee Nov 11 '19 at 11:49

We have that

  • $\sin^2(4x)=16x^2+o(x^2)$

  • $\sin^{-1}(2x)=2x+o(x^2)$


$${\sin^2(4x)}^{\sin^{-1}(2x)}=e^{(2x+o(x^2))\log(16x^2+o(x^2))}\to e^0=1$$

  • $\begingroup$ Thanks for answering! The 2 equations are obtained by maclaurin series right? $\endgroup$ – Scott Lee Nov 11 '19 at 11:45
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    $\begingroup$ Yes of course exactly! $\endgroup$ – user Nov 11 '19 at 11:49
  • $\begingroup$ I really appreciate your help! $\endgroup$ – Scott Lee Nov 11 '19 at 17:36

L'Hospital's rule is not the alpha and omega of limits computations!

It is very simple using equivalents:

We'll first determine the limit of the log: $$\ln \Bigl(\bigl(\sin^2 4x\bigr)^{\arcsin 2x}\Bigr)=2\arcsin 2x\ln(\sin 4x).$$ Now, we have:

  • $\arcsin 2x\sim_0 2x$;
  • $\sin 4x\sim_0 4x$, so $\ln(\sin 4x)\sim_{0^+} \ln(4x)$.

Therefore $\;2\arcsin 2x\ln(\sin 4x)\sim_{0^+} 4x\ln(4x)$, which tends to $0$ by a standard high-school limit. By continuity, the limit of the given expression is equal to $1$.


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