Question

$li{m}_{x\to 1}\frac{\sqrt{1-cos2(x-1)}}{x-1}$

• 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 the left hand limit is not equal to the right hand limit.

Answer 1

The correct option is D Does not exist because the left hand limit is not equal to the right hand limit.

$L.H.L={lim}_{x\to 1^{-}}\frac{\sqrt{1-cos\left[2\right(x-1\left)\right]}}{x-1}={lim}_{x\to 1^{-}}\frac{\sqrt{2si{n}^2(x-1)}}{x-1}=\sqrt{2}.{lim}_{x\to 1^{-}}\frac{\sqrt{si{n}^2(x-1)}}{x-1}=\sqrt{2}{lim}_{x\to 1^{-}}\frac{|sin(x-1)}{x-1}=\sqrt{2}{lim}_{h\to 0}\frac{|sin(-h)|}{-h}=\sqrt{2}{lim}_{h\to 0}\frac{sinh}{-h}=-\sqrt{2}$
Again,
$R.H.L.={lim}_{x\to 1^{+}}\frac{\sqrt{1-cos(x-1)}}{x-1}={lim}_{x\to 1^{+}}\sqrt{2}\frac{|sin(x-1)|}{x-1}$
Put $x=1+h,h>0forx\to 1^{+},h\to 0.={lim}_{h\to 0}\sqrt{2}\frac{|sinh|}{h}={lim}_{h\to 0}\sqrt{2}\frac{sinh}{h}=\sqrt{2}$
L.H.L $\ne$ R.H.L Therefore ${lim}_{x\to 1}f\left(x\right)$ does not exist.