Question

The value of $\underset{x\to -\pi }{lim}\frac{|x+\pi |}{\sin x}$

• A. is equal to $-1$
• B. is equal to $1$
• C. is equal to $\pi$
• D. Does not exits

Answer 1

The correct option is D Does not exits

We have, $f\left(x\right)=\frac{|x+\pi |}{\sin x}$

$\therefore \underset{x\to {\pi }^{-}}{lim}f\left(x\right)=\underset{h\to 0^{+}}{lim}f(-\pi -h)=\underset{h\to 0^{+}}{lim}\frac{|-\pi -h+\pi |}{\sin (-\pi -h)}$

and, $=-\underset{h\to 0^{+}}{lim}\frac{\left|h\right|}{\sin (\pi +h)}=\underset{h\to 0^{+}}{lim}\frac{h}{\sin h}=1$

$\underset{x\to -{\pi }^{+}}{lim}f\left(x\right)=\underset{h\to 0^{+}}{lim}f(-\pi +h)=\underset{h\to 0}{lim}\frac{|-\pi +h+\pi |}{\sin (-\pi +h)}=\underset{h\to 0}{lim}\frac{h}{-\sin h}=-1$

Hence, ${lim}_{x\to -\pi }f\left(x\right)$ does not exist