Limit $\lim_{x \to 1} \ (1-x^2)\tan(\frac{\pi x}{2})$?

Question

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 1

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

Answer 2

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?

Answer 3

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?