Question

Prove that $\underset{y\to x}{lim}\frac{{\cos }^{2}y-{\cos }^{2}x}{y^2-x^2}=-\frac{\sin 2x}{2x}$.

Answer 1

Now,

$\underset{y\to x}{lim}\frac{{\cos }^{2}y-{\cos }^{2}x}{y^2-x^2}$ $\left(\frac{0}{0}\right)$ form

Now applying L'Hospital's rule we get,

$=\underset{y\to x}{lim}\frac{-2\cos y.\sin y}{2y}$

$=-\frac{\sin 2x}{2x}$.