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.


2 Answers 2


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: $$ \lim_{x \rightarrow 0} F(x) = 0 $$

  • $\begingroup$ 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 $\endgroup$ May 24, 2020 at 17:26
  • $\begingroup$ I have now added the explanation to the first limit $\endgroup$ May 24, 2020 at 17:37
  • $\begingroup$ 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 $\endgroup$ May 24, 2020 at 17:44
  • $\begingroup$ 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 $0$. $\endgroup$ May 24, 2020 at 18:01
  • $\begingroup$ @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 $\endgroup$
    – Alex
    May 24, 2020 at 18:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .