Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to solve the limit $$\lim\limits_{t \to \pi/2}\frac{ \int_{\sin t}^{1}e^{x^2\sin t}dx}{\int_{\cos t}^{0}e^{x^2 \cos t}dx}$$

My first method was to try with L'Hopital, i derived using Leibniz rule:

$$\eqalign{\frac{\partial}{\partial t} \left(\int_{\sin t}^{1}e^{x^2\sin t}dx\right)&=\int_{\sin t}^{1}e^{x^2 \sin t}x^2 \cos tdx-e^{\sin ^3 t}\cos t\\ &= \cos t\left(\int_{\sin t}^{1}x^2e^{x^2 \sin t}dx-e^{ \sin ^3 t}\right).\\}$$

In the same manner, we can see that $$\frac{\partial}{\partial t} \left(\int_{\cos t}^{0}e^{x^2 \cos t}dx\right) = -\sin t\left(\int_{\cos t}^{0}x^2e^{x^2 \cos t}dx-e^{\cos ^3 t}\right).$$

So overall we have:

$$\lim\limits_{t \to \pi/2}\frac{ \int_{\sin t}^{1}e^{x^2\sin t}dx}{\int_{\cos t}^{0}e^{x^2 \cos t}dx} = \lim\limits_{t \to { \pi}/{2}} \frac{\cos t\left(\int_{\sin t}^{1}x^2e^{x^2 \sin t}dx-e^{ \sin ^3 t}\right)}{-\sin t\left(\int_{\cos t}^{0}x^2e^{x^2 \cos t}dx-e^{\cos ^3 t}\right)}.$$

But where do we go from here? Could we say that because $\lim\limits_{t \to \pi/2} \frac{\cos t}{\sin t} =0$ then the entire limit goes to $0$? I don't think we can...

Would appreciate any input. Perhaps L'Hopital was not the way.

share|cite|improve this question
If you are familiar with the $erf$ function, the integrals are quite simple. – Claude Leibovici Jun 19 '14 at 14:28
I am not familiar with it, I am supposed to only solve it with tools that were taught in calculus class, and error function was not mentioned. – Oria Gruber Jun 19 '14 at 14:29
Perhaps finding close bounds for $e^x$ around both $1$ and $0$ will allow a sandwich-theorem approach. ...or perhaps would be unnecessarily complicated. – barto Jun 19 '14 at 14:31
up vote 1 down vote accepted

I don't think we can...

Right. Not without looking at the other terms anyway. In particular the factor

$$\int_{\cos t}^0 x^2 e^{x^2\cos t}\,dx - e^{\cos^3 t}$$

in the denominator must be looked at, since that could cancel the vanishing of the numerator.

But, the integral tends to $0$ as $t\to \pi/2$, and the $e^{\cos^3 t}$ tends to $1$, so the denominator does not tend to $0$ as $t\to \pi/2$.

Now a short look to see that the factor in the numerator remains bounded suffices to conclude that the limit is indeed $0$.

Alternatively, we can find out the asymptotic behaviour of numerator and denominator. For that, it is advisable to write $t = \frac{\pi}{2}-u$, and consider the behaviour as $u \to 0$.

The numerator becomes

$$\begin{align} \int_{\sin t}^1 e^{x^2\sin t}\,dx &= \int_{\cos u}^1 e^{x^2\cos u}\,dx\\ &\approx \int_{1-\frac{u^2}{2}}^1 e^{x^2\cos u}\,dx\\ &\approx \frac{u^2}{2}\cdot e^{1}, \end{align}$$

and the denominator

$$\begin{align} \int_{\cos t}^0 e^{x^2\cos t}\,dx &= \int_{\sin u}^0 e^{x^2\sin u}\,dx\\ &\approx \int_u^0 e^{x^2 u}\,dx\\ &\approx -u\cdot e^0. \end{align}$$

So the numerator tends to $0$ quadratically, and the denominator linearly.

share|cite|improve this answer
I was pretty close.. :( – Oria Gruber Jun 19 '14 at 14:35
Indeed. I've added an alternative way to get the result, always good to expand one's toolbox. – Daniel Fischer Jun 19 '14 at 14:38

$$L=\lim_{t \to \pi/2}\frac{ \int_{\sin t}^{1}e^{x^2\sin t}dx}{\int_{\cos t}^{0}e^{x^2 \cos t}dx}$$

$$= \lim_{t \to \pi/2}\frac{ \int_{\sin t}^{1} 1 +(x^2\sin t) +(x^2\sin t)^2/2 +\cdots dx}{\int_{\cos t}^{0} 1 +(x^2\cos t) +(x^2\cos t)^2/2 +\cdots dx}$$

$$= \lim_{t \to \pi/2}\frac{ \left. x +(x^3\sin t)/3 +(x^5\sin ^2t)/10 +\cdots \right|_{x = \sin(t)}^{x=1}}{\left. x +(x^3\cos t)/3 +(x^5\cos ^2t)/10 +\cdots \right|_{x = \cos(t)}^{x=0}}$$

$$= \lim_{t \to \pi/2}\frac{ \left( 1 +(\sin t)/3 +(\sin ^2t)/10 +\cdots \right)-\left( \sin(t) +(\sin^4t)/3 +(\sin ^7t)/10 +\cdots \right)}{-\left( \cos(t) +(\cos^4t)/3 +(\cos ^7t)/10 +\cdots \right)}$$

It is clear that both the numerator and denominator go to $0$. I will now use L'hopitals rule and ignore appropriate terms:

$$= \lim_{t \to \pi/2}\frac{ (\cos t)/3 - \cos(t)}{\sin t}$$

Numerator tends to $0$ and denominator tends to $1$ so we can say $L = 0$.

All ignored terms also tend to $0$ because they contain a term with a cosine.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.