Question

$L=\underset{x\to a}{lim}\frac{|2\sin x-1|}{2\sin x-1}$. Then,

• A. limit does not exist when $a=\frac{\pi }{6}$
• B. $L=-1$ when $a=\pi$
• C. $L=1$ when $a=\frac{\pi }{2}$
• D. $L=1$ when $a=0$

Answer 1

Answer: (A), (B), (C)

limit does not exist when $a=\frac{\pi }{6}$
$L=-1$ when $a=\pi$
$L=1$ when $a=\frac{\pi }{2}$

$L={lim}_{x\to a}\frac{|2sinx-1|}{2sinx-1}$
For $a=\pi /6$,
$L.H.L.=\underset{x\to \frac{{\pi }^{-}}{6}}{lim}\frac{1-2sinx}{2sinx-1}=-1$
$R.H.L=\underset{x\to \frac{{\pi }^{+}}{6}}{lim}\frac{2sinx-1}{2sinx-1}=1$
Hence, the limit does not exist.
For $a=\pi ,{lim}_{x\to \pi }\frac{1-2sinx}{2sinx-1}=-1$ (as in neighborhood of $\pi$, sin x is less than $\frac{1}{2}$).
For $a=\frac{\pi }{2},\underset{x\to \frac{\pi }{2}}{lim}\frac{2sinx-1}{2sinx-1}=1$ (as in neighborhood of $\pi /2$, sin x approaches 1).