# Limit with Integral in it

$$\lim_{x \to 0} \dfrac{\displaystyle \int_0^x \sin \left(\pi t^2/2\right) dt}{x^3}$$ I am having trouble trying to figure out how to compute the limit.

Do I have to take the integral first and then take the limit?

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## 4 Answers

Yes. You will have to take the integral first and then take the limit. However, note that the limit is of the form $\dfrac{0}0$, when you plug in $x=0$. Hence, you can use L'Hospital rule.

Use L'Hospital rule along with Fundamental theorem of Calculus. We then get $$\lim_{x \to 0} \dfrac{\sin(\pi x^2/2)}{3x^2}$$ Now you should be able to finish this off.

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how do you take the integral of sin(x^2)? – user71317 Apr 22 '13 at 18:11
@user71317 There is no need to find the integral of $\sin(x^2)$. All you need is the fact that $$\dfrac{d}{dx} \left(\int_0^x f(t)dt\right) = f(x),$$which is obtained from the fundamental theorem of calculus. – user17762 Apr 22 '13 at 18:12
is it assumed that the limit of the numerator is 0 before using L'Hospitals rule? – user71317 Apr 23 '13 at 4:22
@user71317 It is not assumed but it is zero as $x \to 0$, we have $\int_0^x \sin(\pi t^2/2) dt \to 0$. – user17762 Apr 23 '13 at 5:48

Hint: Use L'Hospital's Rule, using the Fundamental Theorem of Calculus to evaluate the derivative of the top.

Remark: The function $\sin x^2$ does not have an elementary antiderivative, so a strategy based on evaluating the integral will not succeed.

But there is a good alternative to L'Hospital's Rule. Use the power series expansion of $\sin x$ to find the power series expansion of $\sin(\pi t^2/2)$. Then integrate term by term. The answer to your limit problem will pop out.

The first term in the power series expansion of $\sin x$ is $x$. Substitute $\pi t^2/2$, and integrate from $0$ to $x$. We get $\pi x^3/6$. The remaining terms of the power series are negligible in comparison for $x$ near $0$. So the limit of the quotient is $\pi/6$.

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Hint: You can use L'Hospital's Rule

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You also can use $\sin x=x+O(x^3)$ to get the limit. In fact \begin{eqnarray*} \lim_{x\to 0}\frac{\int_0^x\sin(\pi t^2/2)dt}{x^3}= \lim_{x\to 0}\frac{\int_0^x(\pi t^2/2+O(t^6))dt}{x^3}=\lim_{x\to 0}\frac{\pi x^3/6+O(x^7)}{x^3}=\pi/6. \end{eqnarray*}

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