The value of _x0 _0^x^2^2t,dtx x is:

Question

The value of $\displaystyle\lim_{x	o0} \dfrac{\int_0^{x^2}\sec^2t\,dt}{x\sin x}$ is:

Accepted Answer

By L'Hospital's rule: $\lim_{x	o0} \dfrac{\dfrac{d}{dx}\int_0^{x^2}\sec^2t\,dt}{\dfrac{d}{dx}(x\sin x)} = \lim_{x	o0}\dfrac{\sec^2(x^2)\cdot2x}{\sin x+x\cos x}$

$= \lim_{x	o0}\dfrac{2\sec^2(x^2)}{\dfrac{\sin x}{x}+\cos x} = \dfrac{2	imes1}{1+1} = 1$

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