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\lim _{x\to0 }\frac{\int_{0}^{x^{2}}\cos t^{2}dt}{xsinx}=?

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    0

  • Option 4)

    none

 

Answers (1)

As we learnt

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)

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Limit= \lim _{x\to0 }\frac{\frac{d}{dx}\int_{0}^{x^{2}}\cos t^{2}dt}{\frac{d}{dx}\left ( x\sin x \right )}=\lim _{x\to0 }\frac{\cos\left ( x^{2} \right )^{2}.\frac{d\left ( x^{2} \right )}{dx} }{\sin x+x\cos x}

        =\lim _{x\to0 }\frac{2x\cos x^{4}}{\sin x+x\cos x}=\lim _{x\to0 }\frac{2x\cos x^{4}}{x.\left ( \frac{\sin x}{x} \right )+x\cos x}=\lim _{x\to0 }\frac{2\cos x^{4}}{1+\cos x}=1

 

 


Option 1)

1

Option 2)

2

Option 3)

0

Option 4)

none

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