1. ## [SOLVED] Need Help Please

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!

2. Originally Posted by Mr. Engineer
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!
Did you use the distance formula? the two sides of the equilateral triangle that are congruent are $\displaystyle x=\sqrt{(x-0)^2+(1-\cos(x))^2}=\sqrt{x+-2\cos(x)+\sin^2(x)}$?

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 $\displaystyle \frac{3}{4}$

4. Originally Posted by Mr. Engineer
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 $\displaystyle \frac{3}{4}$
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

5. ## Thanks

thanks, thats all i needed to know