A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) \[\frac{3}{2}\]
D) \[-\frac{3}{2}\]
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\tan }^{-1}}x-{{\sin }^{-1}}x}{{{x}^{3}}}\] By L Hospital?s rule, \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{1+{{x}^{2}}}-\frac{1}{\sqrt{1-{{x}^{2}}}}}{3{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-{{x}^{2}}-(1+{{x}^{2}})}}{3{{x}^{2}}(1+{{x}^{2}})\sqrt{1-{{x}^{2}}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{(1-{{x}^{2}})-{{(1+{{x}^{2}})}^{2}}}{3{{x}^{2}}(1+{{x}^{2}})\sqrt{1-{{x}^{2}}}\{\sqrt{1-{{x}^{2}}}+(1+{{x}^{2}})\}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-{{x}^{2}}-x-{{x}^{4}}-2{{x}^{2}}}{3{{x}^{2}}(1+{{x}^{2}})\sqrt{1-{{x}^{2}}}\{\sqrt{1-{{x}^{2}}}+(1+{{x}^{2}})\}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{-3{{x}^{2}}\left( 1+\frac{{{x}^{2}}}{3} \right)}{3{{x}^{2}}(1+{{x}^{2}})\sqrt{1-{{x}^{2}}}\{\sqrt{1-{{x}^{2}}}+(1+{{x}^{2}})\}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{(-1)\left( 1+\frac{{{x}^{2}}}{3} \right)}{(1+{{x}^{2}})\sqrt{1-{{x}^{2}}}\{\sqrt{1-{{x}^{2}}}+1+{{x}^{2}}\}}\] \[=-\frac{1}{2}\] Alternate Method \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\tan }^{-1}}x-{{\sin }^{-1}}x}{{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left[ \begin{align} & \left\{ x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}+...\infty \right\} \\ & -\left\{ x+\frac{1}{2}.\frac{{{x}^{3}}}{3}+\frac{1}{2}.\frac{3}{4}.\frac{{{x}^{5}}}{5}+...\infty \right\} \\ \end{align} \right]}{{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\{-1/3-1/6\}{{x}^{3}}+(1/5-3/40){{x}^{5}}+....\infty }{{{x}^{3}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\left( -\frac{3}{6} \right)+(1/5-3/40){{x}^{2}}+.....\infty =-\frac{1}{2}\]
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