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

Answer: (D)

$2$

Answer 2

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