Question

$\underset{x\to \mathrm{\pi }}{\lim }\frac{\sin x}{x-\mathrm{\pi }}\mathrm{is}$

(a) 1
(b) 2
(c) –1
(d) –2

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ x\to \mathrm{\pi }\end{array}\frac{\sin x}{x-\mathrm{\pi }} \\ \mathrm{put}x-\mathrm{\pi }=h\mathrm{i}.\mathrm{e}.x=\mathrm{\pi }+h\mathrm{where}h\to 0. \\ =\begin{array}{c}\lim \\ h\to 0\end{array}\frac{\sin \left(\mathrm{\pi }+h\right)}{\left(\mathrm{\pi }+h-\mathrm{\pi }\right)} \\ =\begin{array}{c}\lim \\ h\to 0\end{array}\frac{\left(-\sin h\right)}{h} \\ =-\left(\begin{array}{c}\lim \\ h\to 0\end{array}\frac{\sin h}{h}\right)\left[\because \frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{x}\to 01\right] \\ =-1. \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option}\mathrm{C}. \end{gathered}$$