Question

$ \underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin {{x}^{2}}}{1-\cos x} $ is

• A. $ \frac{1}{2} $
• B. $ 0 $
• C. $ 1 $
• D. $ 2 $

Answer 1

The Correct Option is D

$ \underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{sin\,{{x}^{2}}}{1-\cos \,x} $
$=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{sin\,{{x}^{2}}}{2{{\sin }^{2}}\frac{x}{2}} $ $ \left[ \because \,\,1-\cos \,x=\,2{{\sin }^{2}}\frac{x}{2} \right] $
$=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin {{x}^{2}}/{{x}^{2}}}{2{{\sin }^{2}}\left( \frac{x}{2} \right)/{{x}^{2}}} $
$=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin \,{{x}^{2}}}{{{x}^{2}}}+\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\left( \frac{x}{2} \right)}{\frac{{{x}^{2}}}{4}\times 4} $
$=1+\frac{2}{4}\underset{x\to 0}{\mathop{\lim }}\,\,{{\left\{ \frac{\sin \frac{x}{2}}{x/2} \right\}}^{2}}\,\,\,\,\,\,\left[ \because \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin x}{x}=1 \right] $
$=1+\frac{1}{2}\times 1 $
$=2 $