Question

Evaluate $\underset{n\to 0}{lim}\frac{{\sin }^2(x+n)-{\sin }^{2}x}{n}$.

Answer 1

Consider $\frac{{\sin }^2(x+n)-{\sin }^{2}x}{n}$

Using $\sin (A+B)\sin (A-B)={\sin }^{2}A-{\sin }^{2}B$

$=\frac{\sin (x+n+x)\sin (x+n-x)}{n}$

$=\frac{\sin (2x+n)\sin \left(n\right)}{n}$

Now, $\underset{n\to 0}{lim}\frac{\sin (2x+n)\sin \left(n\right)}{n}$

$=\underset{n\to 0}{lim}\sin (2x+n)\times \underset{n\to 0}{lim}\frac{\sin n}{n}$

$=\sin 2x\times 1$ ........... since $\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1$

$=\sin 2x$