# Given $F(x)=\int\limits_x^{x^2}\frac{\sin t}{t}dt$, find $\lim_{x\rightarrow 0}F(x)$ and $\lim_{x\rightarrow 0}F'(x)$

Given $$F(x)=\int\limits_x^{x^2}\frac{\sin t}{t}dt$$, find $$\lim_{x\rightarrow 0}F(x)$$ and $$\lim_{x\rightarrow 0}F'(x)$$

Assume the options are $${-1, 0, 1}$$

Intuitively I'm pretty sure the answer is $$\lim_{x\rightarrow 0}F(x)=0$$ , because $$\frac{\sin t}{t}$$ approaches 1 and we look at a smaller segment as $$x$$ approaches $$0$$, so the area under the graph would approach $$0$$.

Also, I'm pretty sure $$\lim_{x\rightarrow 0}F'(x)=-1$$, because $$x^2 < x$$ for a small enough $$x$$ so we would get minus the value of $$\frac{\sin 0}{0}$$.

I'm having trouble formalizing this, would appreciate help with finding more precise arguments, and also of course let me know if I'm wrong.

Here is a solution for the first limit using only the the fact that $$\lim_{t\rightarrow0}\frac{\sin{t}}{t}=1$$.

For some $$\delta>0$$, $$\Big|\frac{\sin t}{t}\Big|\leq \frac{3}{2}$$ whenever $$|t|<\delta$$. So for $$|x|<\min(\delta,1)$$, $$x^2\leq|x|$$ and $$\left|\int^{x^2}_x\frac{\sin t}{t}\,dt\right|\leq \frac{3}{2}|x-x^2|\leq 3|x|$$ Letting $$x\rightarrow0$$, you obtained that the first limit you are looking for is indeed $$0$$. The second is straight-forward differentiation exercise as pointed out in an early solution.

I will start with the second limit. You can easily evaluate the derivative of $$F$$ as follows: let $$g(t):= \frac{\sin t}{t}$$, and note that $$F(x)=G(x^2)-G(x)$$. Then, also computing the derivative of the composite function, you obtain: $$F'(x)=2x g(x^2) - g(x) = 2x \cdot \frac{\sin x^2}{x^2} - \frac{\sin x}{x}$$ Now, recall that: $$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$ This gives you: $$\lim_{x \rightarrow 0} F'(x) = 2 \cdot 0 \cdot 1 - 1 = -1$$ For the other limit, consider the MacLaurin series of the integrand, and integrate term by term. It is not difficult to see that the result of the limit is $$0$$.

EDIT: we have the Taylor series: $$\frac{\sin t}{t} = \sum_{k=0}^{\infty} \frac{(-1)^n}{(2n+1)!} t^{2n}$$ Now integrate term by term (it is known that this is possible in a case like this one): $$F(x) = \int_{x}^{x^2} \frac{\sin t}{t} dt = \sum_{k=0}^{\infty} \int_{x}^{x^2} \frac{(-1)^n}{(2n+1)!} t^{2n} dt = \sum_{k=0}^{\infty} \frac{(-1)^n}{(2n+1)! (2n+1)} x^{2n+1} (x^{2n+1} - 1)$$ Then, you can see that, as $$x \rightarrow 0$$, $$F(x) \rightarrow 0$$.

EDIT: another solution. Note that $$g(t)=\frac{\sin t}{t}$$ is bounded. To see this, recall that $$|\sin t | \leq |t|$$, which implies that $$|g(t)| \leq 1$$, and thus the function is indeed bounded. Then, we can write: $$\left |\int_{x}^{x^2} \frac{\sin t}{t} dt \right | \leq |x^2 - x| \cdot 1 = |x^2 - x|$$ When $$x \rightarrow 0$$, the RHS goes to $$0$$, and thus: $$\lim_{x \rightarrow 0} F(x) = 0$$

• Thanks i got you on the second limit, care to elaborate about how you use taylor for the first one? I tried and I don't get it at the moment May 24, 2020 at 17:26
• I have now added the explanation to the first limit May 24, 2020 at 17:37
• I see, I've not learned of such taylor developments yet so unfortunately it is not a suitable solution for me. Surely it will benefit other users though so thanks. If you have another idea on how I can prove it I'll be glad to hear May 24, 2020 at 17:44
• I have added another solution; this time, I have proved that the absolute value of the integral is $\leq 0$, which implies that the limit is $0$. May 24, 2020 at 18:01
• @ManuelNorman: why do you have $G(t^2)-G(t)$, not $G(x^2) -G(x)$? Then, I take derivative wrt $x$ on both sides
– Alex
May 24, 2020 at 18:43