Question

$\underset{x\to y}{lim}\frac{{\sin }^{2}x-{\sin }^{2}y}{x^2-y^2}$

• A. $\frac{\sin 2y}{2y}$
• B. $\frac{\cos 2y}{y}$
• C. $\frac{\cos 2y}{2y}$
• D. $\frac{\sin 2y}{y}$

Answer 1

The correct option is A

$\frac{\sin 2y}{2y}$

We have,

$\underset{x\to y}{lim}\frac{{\sin }^{2}x-{\sin }^{2}y}{x^2-y^2}$

This is $\frac{0}{0}$ form.

So, apply L-Hospital rule,

$\underset{x\to y}{lim}\frac{2\sin x\cos x-0}{2x-0}$

$\underset{x\to y}{lim}\frac{\sin 2x}{2x}$

$=\frac{\sin 2y}{2y}$

Hence, this is the answer.