Question

Write the value of $\underset{x\to \mathrm{\pi }}{\lim }\frac{\sin x}{x-\mathrm{\pi }}.$

Answer 1

$$\begin{gathered} \underset{x\to \mathrm{\pi }}{\lim }\left(\frac{\sin x}{x-\mathrm{\pi }}\right) \\ \mathrm{LHL}: \\ \underset{x\to {\mathrm{\pi }}^{-}}{\lim }\left(\frac{\sin x}{x-\mathrm{\pi }}\right) \\ \text{If}x=\mathrm{\pi }-h,\mathrm{then}h\to 0. \\ =\underset{h\to 0}{\lim }\left(\frac{\sin \left(\mathrm{\pi }-h\right)}{\mathrm{\pi }-h-\mathrm{\pi }}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{\sin h}{-h}\right) \\ =-1 \\ \mathrm{RHL}: \\ \underset{x\to {\mathrm{\pi }}^{+}}{\lim }\left(\frac{\sin x}{x-\mathrm{\pi }}\right) \\ \text{If}x=\mathrm{\pi }+h,\mathrm{then}h\to 0. \\ =\underset{h\to 0}{\lim }\left(\frac{\sin \left(\mathrm{\pi }+h\right)}{\mathrm{\pi }+h-\mathrm{\pi }}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{-\sin h}{h}\right) \\ =-1 \\ \therefore \underset{x\to \mathrm{\pi }}{\lim }\left(\frac{\sin x}{x-\mathrm{\pi }}\right)=-1 \end{gathered}$$