$\lim_{x\to 1/2} \frac{\cos(\pi x)}{4x^2-1}$

Question

find the limit : $$\lim_{x\to 1/2} \frac{\cos(\pi x)}{4x^2-1}$$

my try :

$$\lim_{x\to 1/2} \frac{\cos(\pi x)}{4x^2-1}=\lim_{x \to 1/2}\frac{\cos (\pi x-\pi)}{\pi x-\pi} \times \frac{\pi (x-1)}{4x^2-1}\\=\lim_{x\to 1/2} \frac{\cos(\pi x)}{4x^2-1}=\lim_{x \to 1/2}\frac{\cos (\pi x-\pi)}{\pi x-\pi}\times\frac{\pi(x-1)}{(2x-1)(2x+1)}$$

now ??

Answer 1

Let $x=\frac{1}{2}+y$ and use $$\frac{\sin z}{z}\to1$$