# Evaluating $\lim_{x \to \pi/2} (\sin x)^{\tan x}$ using Hôpital's Rule

I am hoping someone can help me check my work here. I need to evaluate this limit:

$$\lim_{x \to \pi/2} (\sin x)^{\tan x}$$

Since $$\sin x$$ and $$\tan x$$ are continuous functions, using the continuity of $$e^x$$, this expression has the equivalent form: $$\lim_{x \to \pi/2} e^{\log{((\sin x)^{\tan x})}}$$

$$\log{(\sin x)^{\tan x}}= \tan x \log{(\sin x)}= \frac{\log{(\sin x)}}{\frac{1}{\tan x}}$$

Taking the limit of this fraction as x goes to $$\pi/2$$ has the indeterminate form of $$-\infty/\infty$$, so we can apply Hôpital's rule.

$$\lim_{x \to \pi/2} \frac{\log{(\sin x)}}{\frac{1}{\tan x}} = \frac{ \frac{1}{\sin x}( - \cos x )}{\frac{-1}{\sin^2 x}}=\frac{ \frac{1}{1} (0) }{\frac{-1}{1}}=0$$

$$\implies \lim_{x \to \pi/2} (\sin x)^{\tan x} = e^{0}=1$$

• Looks fine to me. – Kavi Rama Murthy Nov 17 '19 at 12:27
• Take care when using more than one exponent: $e^{{(\log \sin x)}^{\tan x}}$ without the brackets means $\exp \left( {(\log \sin x}^{\tan x}) \right)$. – Toby Mak Nov 17 '19 at 12:29
• @TobyMak the way you edited the question doesn't look right to me, anyway it should be $e^{\tan x \log( \sin x )}$ which doesn't seem to be what you wrote there. – jeffery_the_wind Nov 17 '19 at 13:20

Your way is correct, as an alternative we can use that

$$(\sin x)^{\tan x}= [(1+(\sin x-1))^{\frac1{\sin x-1}}]^{\tan x(\sin x-1)}\to e^0=1$$

indeed since $$t=\sin x-1 \to 0$$ by standard limits

$$(1+(\sin x-1))^{\frac1{\sin x-1}}=(1+t)^\frac1t \to e$$

and by l'Hospital

$$\lim_{x \to \pi/2}\tan x(\sin x-1)=\lim_{x \to \pi/2}\frac{\sin^2x-\sin x}{\cos x}\stackrel{H.R.}=\lim_{x \to \pi/2}\frac{2\sin x\cos x-\cos x}{-\sin x}=0$$

Setting the limit $$\to0$$ makes things clearer

$$A=\lim_{x \to \pi/2} (\sin x)^{\tan x}=\lim_{y\to0}(\cos y)^{\cot y}$$

$$\ln A=\lim_{y\to0}\dfrac{\ln(\cos y)}{\tan y}$$ which is of the form $$\dfrac00$$

$$\ln A=\lim_{y\to0}\dfrac{-\sin y}{\cos y\sec^2y}=0$$

Just to give a somewhat different approach, note first that since $$\sin x\gt0$$ for $$x$$ near $$\pi/2$$, we have $$\sin x=\sqrt{\tan^2x/(\tan^2x+1)}$$, hence

$$(\sin x)^{\tan x}=\left(1+{1\over\tan^2x}\right)^{-(1/2)\tan x}=\left(\left(1+{1\over\tan^2x}\right)^{\tan^2x}\right)^{-1/(2\tan x)}$$

Next, note that for any $$u\gt0$$ we have

$${1\over e^2}\lt1\lt\left(1+{1\over u}\right)^u\lt e\lt e^2$$

(where the simplifying role of the $$e^2$$'s will soon be apparent). It follows that

$$\left(1\over e\right)^{1/|\tan x|}\lt(\sin x)^{\tan x}\lt e^{1/|\tan x|}$$

Finally, since $$e^{1/|\tan x|}\to1$$ as $$|\tan x|\to\infty$$, the Squeeze Theorem tells us

$$\lim_{x\to\pi/2}(\sin x)^{\tan x}=1$$