Question

$\displaystyle\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$

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

Answer 1

Answer: (D)

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