A) 0
B) 1
C) 2
D) \[\infty \]
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{x{{\sin }^{-1}}x}{\sin {{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{x\left( x+\frac{{{1}^{2}}}{3!}{{x}^{3}}+\frac{{{1}^{2}}{{.3}^{2}}}{5!}{{x}^{5}}+... \right)}{\left( {{x}^{2}}-\frac{{{({{x}^{2}})}^{3}}}{3!}+\frac{{{({{x}^{2}})}^{5}}}{5!}-... \right)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 1+\frac{{{1}^{2}}}{3!}{{x}^{2}}+\frac{{{1}^{2}}{{.3}^{2}}{{x}^{4}}}{5!}+... \right)}{\left( 1-\frac{{{({{x}^{2}})}^{2}}}{3!}+\frac{{{({{x}^{2}})}^{4}}}{5!}-... \right)}\] \[=1\] Alternate Solution: \[\underset{x\to 0}{\mathop{\lim }}\,\frac{x{{\sin }^{-1}}x}{\sin {{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sin {{x}^{2}}}\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}x}{x}\] \[=1\times 1=1\]
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