A) \[\frac{1}{2}\]
B) \[0\]
C) \[1\]
D) \[2\]
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{sin\,{{x}^{2}}}{1-\cos \,x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{sin\,{{x}^{2}}}{2{{\sin }^{2}}\frac{x}{2}}\] \[\left[ \because \,\,1-\cos \,x=\,2{{\sin }^{2}}\frac{x}{2} \right]\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin {{x}^{2}}/{{x}^{2}}}{2{{\sin }^{2}}\left( \frac{x}{2} \right)/{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin \,{{x}^{2}}}{{{x}^{2}}}+\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\left( \frac{x}{2} \right)}{\frac{{{x}^{2}}}{4}\times 4}\] \[=1+\frac{2}{4}\underset{x\to 0}{\mathop{\lim }}\,\,{{\left\{ \frac{\sin \frac{x}{2}}{x/2} \right\}}^{2}}\,\,\,\,\,\,\left[ \because \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin x}{x}=1 \right]\] \[=1+\frac{1}{2}\times 1\] \[=2\]
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