# How to find the following limit? $\lim\limits_{t \to {\pi}/{2}}\frac{ \int_{\sin t}^{1}e^{x^2\sin t}dx}{\int_{\cos t}^{0}e^{x^2 \cos t}dx}.$

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.

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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

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.

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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.

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