Calculate the limit of $\lim_{x\to 1}\frac{1-x^2}{\sin \pi x}$ only with the fundamental limits

Question

I have this limit

$$\lim_{x\to 1}\frac{1-x^2}{\sin \pi x} \tag 1$$

which can be solved, easily, by De l'Hospital's rule (the solution is $2/\pi$). I have considered that $x\to 1$, therefore $t=x-1\to 0$. We have also $t+2=x+1$.

Hence: $$\lim_{x\to 1}\frac{1-x^2}{\sin \pi x} =\lim_{t\to 0}\frac{-t(t+2)}{\sin \pi (t+1)}=\lim_{t\to 0}\frac{\pi (t+1)}{\sin \pi (t+1)}\cdot \frac{-t(t+2)}{\pi (t+1)}\tag 2$$ but with the

$$\lim_{t\to 0}\frac{\pi (t+1)}{\sin \pi (t+1)}\cdot \frac{-t(t+2)}{\pi (t+1)}\equiv 0$$ and I have not found the solution $2/\pi$. Why?

Answer 1

Let $t=x-1$. Then we have

$$\frac{1-x^2}{\sin(\pi x)}=\frac{-t(t+2)}{\sin(\pi (t+1))}=\frac{t(t+2)}{\sin(\pi t)}$$

Now, set $\pi t=y$ to find

$$\lim_{x\to 1}\frac{1-x^2}{\sin(\pi x)}=\lim_{y\to 0}\left(\frac1\pi(y/\pi+2)\frac{y}{\sin(y)}\right)$$

Can you finish?

Answer 2

Your initial change of variables was good, but then you have to use the correct manipulation:

$$\lim_{t\to 0} \frac{-t}{\sin(\pi t + \pi)} \cdot (t+2) = \lim_{t\to 0} \frac{-\pi t}{-\sin(\pi t)} \cdot \frac{t+2}{\pi} = 1 \cdot \frac{2}{\pi} = \frac{2}{\pi}$$

Answer 3

I'll propose a variant to simplify a bit the computations: setting $x=1-t$, which is the same as $t=1-x$, you get $$\frac{1-x^2}{\sin(\pi x)}=\frac{(1-x)(1+x)}{\sin(\pi x)}=\frac {t(2-t)}{\sin(\pi-\pi t)}=\underset{\stackrel{\textstyle\downarrow}{\tfrac1\pi}}{\frac{t}{\sin\pi t}}\,\underset{\stackrel{\textstyle\downarrow}{2}}{(2-t)}$$