Question

Solve :
$\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin (x-h)}{h}$

Answer 1

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

$={lim}_{h\to 0}\frac{2\cos \left(\frac{x+h+x-h}{2}\right)\sin \left(\frac{x+h-x+h}{2}\right)}{h}$ using transformation angle formula $\sin C-\sin D=2\sin \left(\frac{C-D}{2}\right)\cos \left(\frac{C+D}{2}\right)$

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

$={lim}_{h\to 0}\frac{\sin h}{h}\times 2\cos x$

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

$=2\cos x$