I'm having trouble solving $\lim_{x \to 0} \frac{1}{x^2} \int_0^x \frac{t + t^2}{1 + \sin(t)} dt$ (Pset 3, #8 of UCLA math gre practice problems)

Question

Normally, I would solve the integral and then evaluate the limit, but the integral seems particularly difficult, and I'm wondering if there's some sort of trick involved. The solution said to just use L'Hospital and the Fundamental Theorem of Calculus, but I don't know where to begin. I've tried moving the $\frac{1}{x^2}$ into the integral, but apart from the fact that I don't know if that's allowed, it didn't seem to help...

Answer 1

Hint: Set $f(x)=\int_0^x\frac{t+t^2}{1+\sin t}\,\mathrm dt$. You have to find the limit of $\;\frac{f(x)}{x^2}$ as $x\to 0$. Apply L'Hospital's rule, after you've calculated $f'(x)$ by the first fundamental theorem of integral calculus.

Answer 2

You certainly can move the $1/x^2$ inside the integral but it won't help.

Consider that for any $\epsilon\in (0,1)$ there exists $x\in (0,1)$ such that for all $|t|\le x$ we have $1+\sin t=(1+t)(1+f(t))$ where $1-\epsilon<1+f(t)<1+\epsilon.$ This bounds $x^{-2}\int_0^x \frac {t+t^2}{1+\sin t}dt$ between $x^{-2}\int_0^x \frac {t+t^2}{(1+t)(1\pm \epsilon)}dt=$ $x^{-2}\int_0^x\frac {t}{1\pm \epsilon}dt=$ $\frac {1}{2}\cdot \frac {1}{1\pm \epsilon}.$

Answer 3

You can use the substitution $z=t^2$ so that $dt=\frac{dz}{2\sqrt{z}}$ and the desired limit is $$\frac{1}{2}\lim_{x\to 0}\frac{1}{x^2}\int_{0}^{x^2}\frac{1+\sqrt{z}}{1+\sin\sqrt{z}}\,dz$$ The integrand tends to $1$ as $z\to 0$ and hence by Fundamental Theorem of Calculus the expression under limit above tends to $1$ and the desired limit is $1/2$.

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Alternatively you can also use substitution $t=xz$ so that expression under limit is transformed into $$\int_{0}^{1}\frac{z+xz^2}{1+\sin (xz) } \, dz$$ The integrand is continuous as a function of $x, z$ and hence the limit can be taken inside the integral to give the desired limit as $\int_{0}^{1}z\,dz=1/2$.

Answer 4

Let $$F(x) = \int^x_0 \frac{t + t^2}{1 + \sin t} dt.$$ Near $x = 0$, this function has continuous derivatives up to and including the third degree and beyond. Therefore, by Taylor's Theorem at the third order, there exists a continuous function $G(x)$ such that $$F(x) = F(0) + x F'(0) + \frac{x^2}{2} F''(0) + x^3 G(x).$$ But $$F'(x) = \frac{x + x^2}{1 + \sin x}.$$ Therefore $$F(0) = 0, \quad F'(0) = 0,$$ and $$F''(0) = \left(\frac{d}{dx}\left(\frac{x + x^2}{1 + \sin{x}}\right)\right)_{x=0} = \left(\frac{1 + 2x}{1 + \sin{x}} - \frac{x + x^2}{1 + \sin{x}}\frac{\cos{x}}{1 + \sin{x}}\right)_{x=0} = 1,$$ so that $$F(x) = \frac{x^2}{2} + x^3 G(x).$$

Upon substitution, we get: $$\lim_{x→0}\frac{1}{x^2}\int^x_0\frac{t + t^2}{1 + \sin{t}}dt = \lim_{x→0}\frac{F(x)}{x^2} = \lim_{x→0}\left(\frac{1}{2} + xG(x)\right) = \frac{1}{2} + 0 G(0) = \frac{1}{2},$$ since $G(x)$ is continuous at $x = 0$.