Find the limit as $\displaystyle x$ appraoches zero of the ratio of the triangle's area to the total shaded area of the figure.

See the attached picture.

Thank You!

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- May 1st 2008, 03:26 PM #1

- May 1st 2008, 03:31 PM #2

- May 1st 2008, 04:58 PM #3
## So far

I used the typical area formula for the triangle, then for the area of the shaded region, i constructed a rectangle, then subtracted they two triangles it formed, leaving the area of the shaded region as an integral. the hard part is the actual ratio, which is

$\displaystyle Ratio=\frac{x-xcos(x)}{2sin(x)-2xcos(x)}$

Then one of the steps the author of my book took was to take the derivative of the numerator and the denominator separately, so it looks like this...

$\displaystyle =\frac{1+xsin(x)-cos(x)}{2xsin(x)}$

Don't worry about the steps to get to this because this is actually where i am having my trouble...how do you justify taking the derivative of the numerator and denominator separately and what do you do next. The answer is $\displaystyle \frac{3}{4}$

- May 1st 2008, 05:10 PM #4
This is L'hopital's rule which states that if a lim $\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=\frac{0}{0},\frac{ \infty}{\infty}$

THen $\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=\lim_{x\to{c}}\fra c{f'(x)}{g'(x)}$

Then you will have to use L'hopital's again because your new thing yields 0/0

So you haev $\displaystyle \lim_{x\to{0}}\frac{1+x\sin(x)-\cos(x)}{2x\sin(x)}=\lim_{x\to{0}}\frac{x\cos(x)+\ sin(x)+\sin(x)}{2x\cos(x)+2\sin(x)}=\frac{0}{0}$

Differentiating again will yield the desired result

- May 1st 2008, 06:42 PM #5