Limit $\lim_{x \to 1} \ (1-x^2)\tan(\frac{\pi x}{2})$? Somebody, help me please with this limit
$$\lim_{x \to 1} \ (1-x^2)\tan\left(\frac{\pi x}{2}\right)$$
I've tried to dissasemble this equation into
$$\lim_{x \to 1} \ (1-x^2)\frac{\sin\frac{\pi x}{2}}{\cos\frac{\pi x}{2}}$$
Sinus will be equal to 1, so:
$$\lim_{x \to 1} \ \frac{(1-x^2)}{\cos\frac{\pi x}{2}}$$
And then I don't know what should i do next.
I feel very sorry that i didn't mention it before, but i need to solve it without l'hospital rule.
 A: Decompose also $1-x^2$:
$$
\lim_{x\to1}(1-x^2)\tan\frac{\pi x}{2}=
\lim_{x\to1}\frac{1-x}{\cos(\pi x/2)}(1+x)\sin\frac{\pi x}{2}
$$
The second and third factors have limits $2$ and $1$ respectively; for the first fraction, consider instead
$$
\lim_{x\to1}\frac{\cos(\pi x/2)}{x-1}=
\lim_{x\to1}\frac{\cos(\pi x/2)-\cos(\pi/2)}{x-1}
$$
which is the derivative at $1$ of the function $f(x)=\cos(\pi x/2)$; since
$$
f'(x)=-\frac{\pi}{2}\sin\frac{\pi x}{2}
$$
and so $f'(1)=-\pi/2$, your limit is
$$
-\frac{1}{-\pi/2}\cdot2\cdot1=\frac{4}{\pi}
$$
If using the derivative is not allowed, then use a substitution: for
$$
\lim_{x\to1}\frac{1-x}{\cos(\pi x/2)}
$$
set $1-x=2t/\pi$ so $x=1-2t/\pi$ and
$$
\frac{\pi}{2}x=\frac{\pi}{2}-t
$$
so the limit becomes
$$
\lim_{t\to0}\frac{2t/\pi}{\cos(\pi/2-t)}=
\lim_{t\to0}\frac{2}{\pi}\frac{\sin t}{t}
$$
A: OK, this is not sleek. After you've got the last expression write the denominator as $1 -  \frac{\pi^2 x^2}{4} + O(x^4)$. Then use L'Hospital's rule. What do you get? 
A: Use De L'Hospital!
$$\lim_{x\to 1} \frac{1-x^2}{\cos (\frac{\pi x}{2})}=\lim_{x\to 1} \frac{4x}{\pi\sin (\frac{\pi x}{2})}$$
Can you take it from here?
