Question

$\underset{x\to 1}{lim}\frac{\sqrt{1-\cos 2(x-1)}}{x-1}$ is equal to?

• A. Exists and it equal $\sqrt{2}$
• B. Exists and it equal $-\sqrt{2}$
• C. Does not exists because $(x-1)\to 0$
• D. Does not exists because left hand limit is not equal to right hand limit

Answer 1

Answer: (D)

Does not exists because left hand limit is not equal to right hand limit

$\underset{x\to 1}{lim}\frac{\sqrt{1-\cos 2(x-1)}}{x-1}=\underset{x\to 1}{lim}\frac{\sqrt{2{\sin }^2(x-1)}}{x-1}=\underset{x\to 1}{lim}\frac{\sqrt{2}\left|\sin (x-1)\right|}{(x-1)}$

$L.H.L=\underset{x\to 1^{-}}{lim}\frac{\sqrt{2}\left|\sin (x-1)\right|}{(x-1)}=\underset{h\to 0}{lim}-\frac{\sqrt{2}\left|\sin (1-h+1)\right|}{(1-h+1)}$

$=\underset{h\to 0}{lim}\frac{\sqrt{2}|-\sin h|}{-h}=-\sqrt{2}\underset{h\to 0}{lim}\frac{\sin h}{h}=-\sqrt{2}.1=-\sqrt{2}$

$R.H.L=\underset{x\to 1^{+}}{lim}\frac{\sqrt{2}\left|\sin (x-1)\right|}{(x-1)}=\underset{h\to 0}{lim}\frac{\sqrt{2}\left|\sin (1+h-1)\right|}{(1+h-1)}$

$=\underset{h\to 0}{lim}\frac{\sqrt{2}\left|\sin h\right|}{h}=\sqrt{2}\underset{h\to 0}{lim}\frac{\sin h}{h}=\sqrt{2}.1=\sqrt{2}$

Since $L.H.L\ne R.H.L$

$\therefore \underset{x\to 1}{lim}\frac{\sqrt{1-\cos 2(x-1)}}{x-1}$ does not exists.