Limit proof

**Siron** · July 18th 2011, 05:40 AM

Hi, I've to proof that:

$\lim_{ x \to 0} \int_{x}^{\frac{\pi}{2}} \frac{1-\cos(u)}{\sin^2(u)}du=1$

This is what I found:

$\lim_{ x \to 0} \int_{x}^{\frac{\pi}{2}} \frac{1-\cos(u)}{\sin^2(u)}du=\lim_{x \to 0} \left(\int_{x}^{\frac{\pi}{2}} \frac{du}{\sin^2(u)} - \int_{x}^{\frac{\pi}{2}}\frac{\cos(u)du}{\sin^2(u) }\right)$
$=\lim_{x \to 0} \left( \left[-\cot(u)\right]_{x}^{\frac{\pi}{2}}+\left[\frac{1}{\sin(u)}\right]_{x}^{\frac{\pi}{2}}\right)$
$=\lim_{x \to 0} \left(\cot(x)+1-\frac{1}{\sin(x)}\right)$

The problem is here, because $\lim_{x \to 0} \cot(x)=+\infty$ and $\lim_{x \to 0} \frac{1}{\sin(x)}=\infty}$ I don't now how to go further.

Thanks in advance!

**Sudharaka** · July 18th 2011, 05:53 AM

Originally Posted by Siron

Hi, I've to proof that:

$\lim_{ x \to 0} \int_{x}^{\frac{\pi}{2}} \frac{1-\cos(u)}{\sin^2(u)}du=1$

This is what I found:

$\lim_{ x \to 0} \int_{x}^{\frac{\pi}{2}} \frac{1-\cos(u)}{\sin^2(u)}du=\lim_{x \to 0} \left(\int_{x}^{\frac{\pi}{2}} \frac{du}{\sin^2(u)} - \int_{x}^{\frac{\pi}{2}}\frac{\cos(u)du}{\sin^2(u) }\right)$
$=\lim_{x \to 0} \left( \left[-\cot(u)\right]_{x}^{\frac{\pi}{2}}+\left[\frac{1}{\sin(u)}\right]_{x}^{\frac{\pi}{2}}\right)$
$=\lim_{x \to 0} \left(\cot(x)+1-\frac{1}{\sin(x)}\right)$

The problem is here, because $\lim_{x \to 0} \cot(x)=+\infty$ and $\lim_{x \to 0} \frac{1}{\sin(x)}=\infty}$ I don't now how to go further.

Thanks in advance!

Hi Siron,

$\lim_{x \to 0} \left(\cot(x)+1-\frac{1}{\sin(x)}\right)=\lim_{x \to 0} \left(\frac{\cos x+\sin x-1}{\sin x}\right)$

Since this limit is in a indeterminate form I suggest you use the L'Hopital's rule.

**Siron** · July 18th 2011, 06:03 AM

Thanks!
I've solved it now.

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