1. ## triangle limit

Here's a cool little limit problem you may wish to tackle.

If $\displaystyle T({\theta})$ is the area of the right triangle ABC, and $\displaystyle S({\theta})$ is the area of the segment formed by chord AB and the arc of the unit circle subtended by the central angle theta.

Find $\displaystyle \lim_{\theta\rightarrow{0^{+}}}\frac{T({\theta})}{ S({\theta})}$

2. Originally Posted by galactus
Here's a cool little limit problem you may wish to tackle.

If $\displaystyle T({\theta})$ is the area of the right triangle ABC, and $\displaystyle S({\theta})$ is the area of the segment formed by chord AB and the arc of the unit circle subtended by the central angle theta.

Find $\displaystyle \lim_{\theta\rightarrow{0^{+}}}\frac{T({\theta})}{ S({\theta})}$
Hello,

in my opinion the limit must be zero:

Area of triangle ABC: $\displaystyle a_{ABC} = \frac{1}{2}(1-\cos(\theta)) \cdot \sin(\theta)$

Area of the segment: $\displaystyle a_{seg} = \frac{\theta}{360} \cdot \pi \cdot 1^2 - \frac{1}{2}\sin(\theta) \cdot 1^2$

$\displaystyle \lim_{\theta~\rightarrow~0} \left( \frac{\frac{1}{2}(1-\cos(\theta)) \cdot \sin(\theta)}{ \frac{\theta}{360} \cdot \pi - \frac{1}{2}\sin(\theta) }\right)$ = $\displaystyle \lim_{\theta~\rightarrow~0} \left( \frac{180 \cdot (1-\cos(\theta)) \cdot \sin(\theta)}{ \theta \cdot \pi - 180 \cdot \sin(\theta) }\right)$ = $\displaystyle \lim_{\theta~\rightarrow~0} \left( \frac{180 \cdot (1-\cos(\theta))}{ \frac{\theta \cdot \pi}{\sin(\theta)} - 180 }\right)$

Since $\displaystyle \lim_{\theta~\rightarrow~0}(1-\cos(\theta) = 0$ and

$\displaystyle \lim_{\theta~\rightarrow~0}\left(\frac{\theta}{\si n(\theta)} \right)=1$ the numerator of the fraction will become zero and the denominator something ugly like $\displaystyle \pi - 180$ but it will not be zero.

Thus the complete fraction must be zero.

3. Earboth, are you sure you used radians. You wrote $\displaystyle \frac{\theta}{360}$ maybe it should be $\displaystyle \frac{\theta}{2\pi}$.

4. Originally Posted by ThePerfectHacker
Earboth, are you sure you used radians. You wrote $\displaystyle \frac{\theta}{360}$ maybe it should be $\displaystyle \frac{\theta}{2\pi}$.
Hello,

no, I used degree. In my opinion it doesn't matter whether you use radians or degree.

5. Area of triangle:

$\displaystyle T({\theta})=\frac{1}{2}sin{\theta}(1-cos{\theta})$

Area of circular segment is given by $\displaystyle \frac{1}{2}r^{2}({\theta}-sin{\theta})$. Our radius = 1.

$\displaystyle S({\theta})=\frac{1}{2}({\theta}-sin{\theta})$

Therefore, we have:

$\displaystyle \lim_{\theta\rightarrow{0}^{+}}\frac{T({\theta})}{ S({\theta})}=\lim_{\theta\rightarrow{0}^{+}}\frac{ sin{\theta}-sin{\theta}cos{\theta}}{{\theta}-sin{\theta}}$

You can use L'Hopital:

$\displaystyle \lim_{\theta\rightarrow{0}^{+}}\frac{cos{\theta}+s in^{2}{\theta}-cos^{2}{\theta}}{1-cos{\theta}}$

identity: $\displaystyle sin^{2}{\theta}-cos^{2}{\theta}=-cos(2{\theta}):$

$\displaystyle =\lim_{\theta\rightarrow{0}^{+}}\frac{cos{\theta}-cos({2\theta})}{1-cos{\theta}}$

L'Hopital again:

$\displaystyle =\lim_{\theta\rightarrow{0}^{+}}\frac{2sin(2{\thet a})-sin{\theta}}{sin{\theta}}$

L'Hopital again:

$\displaystyle =\lim_{\theta\rightarrow{0}^{+}}\frac{4cos(2{\thet a})-cos{\theta}}{cos{\theta}}=\boxed{3}$

Of course, you can manipulate it without L'Hopital. Whatever you want to do.