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.

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

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

  • $\begingroup$ Looks fine to me. $\endgroup$ – Kavi Rama Murthy Nov 17 '19 at 12:27
  • $\begingroup$ Take care when using more than one exponent: $e^{{(\log \sin x)}^{\tan x}}$ without the brackets means $\exp \left( {(\log \sin x}^{\tan x}) \right)$. $\endgroup$ – Toby Mak Nov 17 '19 at 12:29
  • $\begingroup$ @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. $\endgroup$ – 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.