Evaluate $\lim\limits_{x\to 0}\frac{\int\limits_0^{x^2}\sec^2 t \, dt}{x\sin x}$

Question

Question :
Evaluate $\lim\limits_{x\to 0}\dfrac{\int\limits_0^{x^2}\sec^2 t \, dt}{x\sin x}$.

I hahe not got any idea how to evaluate. How can I do this?

Answer 1

You may use $\int \sec^2 x\, dx = \tan x$:

$$\dfrac{\int\limits_0^{x^2}\sec^2 t \, dt}{x\sin x} = \frac{\tan (x^2)}{x \sin x} = \frac{\sin (x^2)}{x^2}\cdot\frac{x}{\sin x}\cdot \frac{1}{\cos (x^2)} \stackrel{x \rightarrow 0}{\longrightarrow}1$$

Answer 2

When you are walking very close to $x=0$ (both sides I mean), you will find out that $$\tan(x^2)\approx x^2, ~~\sin(x)\approx x$$ Use these two facts. I assumed you can handle the definite integral first.

Answer 3

Using L'Hospital rule $$y=\dfrac{\int\limits_0^{x^2}\sec^2 (t) \, dt}{x\sin x}=\frac{\frac d {dx}\left({\int\limits_0^{x^2}\sec^2 (t) \, dt}\right)}{\sin(x)+x \cos(x)}$$ Now, for the numerator, use the fundamental theorem of calculus to get $$y=\frac{2x \sec^2 (x^2) }{\sin(x)+x \cos(x)}$$ Close to $x=0$, the denominator is $\sim 2x$ and the remaining looks to be simple.

Answer 4

This is preferable when the integrand can't be integrated easily. By virtue of the standard limit $\lim_{x\to 0}\dfrac{\sin x} {x} =1$ the denominator $x\sin x$ can be replaced with $x^2$ and then putting $z=x^2$ we can see that the desired limit is equal to $$\lim_{z\to 0}\frac{1}{z}\int_{0}^{z}\sec^2 t\, dt$$ By fundamental theorem of calculus the above limit is equal to $\sec^2 0=1$.