# Find $\lim\limits_{x\to\frac\pi2}(1-\sin x)\tan^2 x.$

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

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

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