Question

${lim}_{x\to 0}\frac{sin(a+x)+sin(a-x)-2sina}{xsinx}$

Answer 1

We have,

${lim}_{x\to 0}\frac{\sin (a+x)-\sin (a-x)-2\sin a}{x\sin x}$

$={lim}_{x\to 0}\frac{2\sin \frac{(a+x+a-x)}{2}\cos \frac{(a+x-a+x)}{2}-2\sin a}{x\sin x}$

$={lim}_{x\to 0}\frac{2\sin \frac{2a}{2}\cos \frac{\left(2x\right)}{2}-2\sin a}{x\sin x}$

$={lim}_{x\to 0}\frac{2\sin a\cos x-2\sin a}{x\sin x}$

$={lim}_{x\to 0}\frac{2\sin a(\cos x-1)}{x\sin x}$

$={lim}_{x\to 0}\frac{2\sin a(1-2{\sin }^2\frac{x}{2}-1)}{x\sin x}$

$={lim}_{x\to 0}\frac{-4\sin a{\sin }^2\frac{x}{2}}{x\sin x}$

$={lim}_{x\to 0}\frac{-4\sin a{\sin }^2\frac{x}{2}}{x2\sin \frac{x}{2}\cos \frac{x}{2}}$

$={lim}_{x\to 0}\frac{-2\sin a\sin \frac{x}{2}}{x\cos \frac{x}{2}}\frac{0}{0}form$

Applying L’ Hospital rule and we get,

$={lim}_{x\to 0}\frac{-2\sin a\cos \frac{x}{2}\times \left(\frac{1}{2}\right)}{x(-\sin \frac{x}{2})\frac{1}{2}+\cos \frac{x}{2}}$

$={lim}_{x\to 0}\frac{-\sin a\cos \frac{x}{2}}{\frac{x}{2}(-\sin \frac{x}{2})+\cos \frac{x}{2}}$

Taking limit and we get,

$=\frac{-\sin a\cos 0}{0(-\sin 0)+\cos 0}$

$=\frac{-\sin a}{0+1}$

$=-\sin a$

Hence, this is the answer.