1
$\begingroup$

I'm stuck finding following limit.

$$\lim_{x\to\frac\pi2}(1-\sin x)\tan^2 x.$$

Attempt: I tried to use L'Hopital rule but I cant find the solution

$\endgroup$
2
  • $\begingroup$ i think the searched limit is $\frac{1}{2}$ $\endgroup$ – Dr. Sonnhard Graubner Nov 12 '15 at 18:33
  • 1
    $\begingroup$ i tried to use L' Hospital rule but i cant find the solution $\endgroup$ – Raio Nov 12 '15 at 18:34
6
$\begingroup$

HINT :

$$\tan^2x=\frac{\sin^2x}{\cos^2x}$$ and $$\cos^2x=1-\sin^2x=(1-\sin x)(1+\sin x)$$

$\endgroup$
2
$\begingroup$

I usually advise to “go at $0$”: do the substitution $x=\pi/2-t$, so the limit becomes $$ \lim_{t\to0}(1-\cos t)\cot^2t= \lim_{t\to0}\frac{1-\cos t}{\sin^2t}\cos^2t= \lim_{t\to0}\frac{1-\cos t}{t^2}\frac{t^2}{\sin^2t}\cos^2t $$

If you want to use l'Hôpital, do $$ \lim_{x\to\pi/2}\frac{1-\sin x}{\cot^2x} \overset{\mathrm{(H)}}{=} \lim_{x\to\pi/2}\frac{-\cos x}{-2\cot x/\sin^2x} $$ Now simplify.

$\endgroup$

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.