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\lim_{x=0} \frac{\int_{0}^{x^{2}}{(\tan^{-1}t)^{2} dt= }}{\int_{0}^{x^{4}}{(\sin\sqrt{t} dt})} is equal to

  • Option 1)

    1

  • Option 2)

    -1

  • Option 3)

    -\frac{1}{2}

  • Option 4)

    \frac{1}{2}

 

Answers (1)

best_answer

 

NEWTON LEIBNITZ THEOREM -

\frac {d}{dt}\left ( \int_{f(t)}^{\phi (t))}F(x)dx \right )=F(\phi(t))\phi^{'}(t)-F(f(t))f^{'}(t)

-

 

 \lim_{x\rightarrow 0}\frac{\int_{0}^{x^{2}}(\tan ^{-1}t)^{2}dt}{\int_{0}^{x^{4}}(\sin \sqrt{t})dt}

\left [ \because \int_{\alpha }^{\beta }f(x)dx=f(\beta )\frac{d\beta }{dx}-f(\alpha )\frac{d\alpha }{dx} \right ]

=>\lim_{x\rightarrow 0}\frac{(\tan ^{-1}x^{2})^{2}\times 2x}{(\sin \sqrt{x^{4}})\times 4x^{3}}

\lim_{x\rightarrow 0}\frac{(\tan ^{-1}x^{2})^{2}}{\sin x^{2}}\times \frac{1}{2x^{2}}

=>\frac{1}{2}\lim_{x\rightarrow 0} \frac{(\tan ^{-1}x^{2})^{2}}{\frac{\sin x^{2}}{x^{2}}\times x^{4}}

=>\lim\frac{1}{2}_{x\rightarrow 0}\times \left ( \frac{\tan ^{-1}x^{2}}{x^{2}} \right )^{2}\times \frac{x^{2}}{\sin x^{2}}

=\frac{1}{2}\times 1\times 1

=\frac{1}{2}


Option 1)

1

Option is incorrect

Option 2)

-1

Option is incorrect

Option 3)

-\frac{1}{2}

Option is incorrect

Option 4)

\frac{1}{2}

Option is correct

Posted by

prateek

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