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.

Answer

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
$$
\lim_{t\to0}\frac{2t/\pi}{\cos(\pi/2-t)}=
\lim_{t\to0}\frac{2}{\pi}\frac{\sin t}{t}
$$