Question

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

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

Answer 1

Answer: (D)

$\underset{x\to 1}{\lim }\frac{\sqrt{1-\cos 2(x-1)}}{x-1}$
$=\underset{x\to 1}{\lim }\frac{\sqrt{2{\sin }^2(x-1)}}{x-1}$
$=\underset{x\to 1}{\lim }\frac{\sqrt{2}|\sin (x-1)|}{(x-1)}$
$LHL=\underset{x\to 1^{-}}{\lim }\frac{\sqrt{2}|\sin (x-1)|}{(x-1)}$
$=\underset{h\to 0}{\lim }-\frac{\sqrt{2}|\sin (1-h-1)|}{(1-h-1)}$
$=\underset{h\to 0}{\lim }\frac{\sqrt{2}|-\sin h|}{-h}$
$=-\sqrt{2}\underset{h\to 0}{\lim }\frac{\sin h}{h}$
$=-\sqrt{2}\cdot 1=-\sqrt{2}$
$RHL=\underset{x\to 1^{+}}{\lim }\frac{\sqrt{2}|\sin (x-1)|}{(x-1)}$
$=\underset{h\to 0}{\lim }\frac{\sqrt{2}|\sin (1+h-1)|}{(1+h-1)}$
$=\underset{h\to 0}{\lim }\frac{\sqrt{2}|\sin h|}{h}$
$=\sqrt{2}\underset{h\to 0}{\lim }\frac{\sin h}{h}$
$=\sqrt{2}\cdot 1=\sqrt{2}$
Since LHL $\ne$ RHL,
$\therefore \underset{x\to 1}{\lim }\frac{\sqrt{1-\cos 2(x-1)}}{x-1}$ does not exist.