Question

Evaluate : ${lim}_{x\to 0}\frac{\sin 2x}{5x}$

Answer 1

$\begin{array}{c}Wehave\\ {lim}_{x\to 0}\frac{\sin \left(2x\right)}{5x}\\ =\frac{1}{5}{lim}_{x\to 0}\frac{\sin \left(2x\right)}{x}\\ Now,byu\sin g{L}^{\prime }HospitalRule\\ {lim}_{x\to 0}\frac{\sin \left(2x\right)}{x}={lim}_{x\to 0}\frac{\frac{d}{dx}\left[\sin 2x\right]}{\frac{d\left(x\right)}{dx}}\\ =\frac{1}{5}{lim}_{x\to 0}\frac{\cos \left(2x\right)\frac{d\left(2x\right)}{dx}}{\frac{d\left(x\right)}{dx}}\\ =\frac{1}{5}{lim}_{x\to 0}\frac{2.\cos \left(2x\right)\frac{d\left(x\right)}{dx}}{\frac{d\left(x\right)}{dx}}\\ =\frac{1}{5}{lim}_{x\to 0}\frac{2\cos \left(2x\right).1}{\frac{d\left(x\right)}{dx}}\left(powerrule\right)\\ =\frac{1}{5}.\frac{\cos \left(2{lim}_{x\to 0}x\right)}{{lim}_{x\to 0}1}\\ =\frac{1}{5}.\frac{2\cos \left(2.0\right)}{1}\\ =\frac{1}{5}.\frac{2\cos 0}{1}\\ =\frac{1}{5}\times 2\\ =\frac{2}{5}\\ Hence,itistherequiredanswer.\end{array}$