have trouble with this limit question

Question

a) By considering the areas of the triangle OAD, the sector OAC and the triangle OBC, show that $(\cos \theta)(\sin \theta) < \theta < \frac{\sin\theta}{\cos\theta}$ I find out: Area of OAD=$\frac{1}{2}OD\cdot AD \cdot \sin \theta$ Area of OAC=$\frac{1}{2}OC^2 \theta$ Area of OBC=$\frac{1}{2}OC\cdot BC \cdot\sin\theta$ Now I'm stuck at how to apply this to prove How to prove?

(b) Use (a) and the Squeeze Theorem to show that $\displaystyle\lim_{\theta\to 0^+}\frac{\sin\theta}{\theta}= 1$

Answer 1

Hint: WORK IN RADIANS! $a)$ $$\text{Area of }\Delta OAD=\dfrac{1}{2}\cdot OD\cdot AD\\ \text{Area of sector }OAC=\dfrac{\theta}{360}\pi (OA)^2\\ \text{Area of }\Delta OBC=\dfrac{1}{2}\cdot OC\cdot BC$$ See that $$\text{Area of }\Delta OAD<\text{Area of sector }OAC<\text{Area of }\Delta OBC\\ \implies \dfrac{1}{2}\cdot \cos\theta\cdot \sin\theta<\dfrac{\theta}{360}\pi (OA)^2<\dfrac{1}{2}\cdot 1\cdot BC$$ Now, $$DC=1-\cos\theta\\ BC=\tan\theta$$

$b)$ Then, after doing $a)$, use the fact that $$\dfrac{1}{2}\sin\theta\cos\theta<\theta/2\\ \theta/2<\dfrac{1}{2}\tan\theta$$ Then use the squeeze theorem. The limit follows.

Answer 2

The geometry has been discussed in comments. So we have $$\frac{1}{2}\cos\theta\lt \frac{1}{2}\theta.$$ Multiply through by $2$. We get $\cos\theta\sin\theta\lt \theta$.

Divide both sides by $\theta\cos\theta$. We get $$\frac{\sin\theta}{\theta}\lt \frac{1}{\cos\theta}.\tag{1}$$

We also got from the geometry that $$\frac{1}{2}\theta\lt \frac{1}{2}\frac{\sin \theta}{\cos\theta}.$$ Multiplying by $2$, and rearranging, we get $\theta\cos\theta\lt \sin\theta$. Divide through by $\theta$. We get $$\cos\theta\lt\frac{\sin\theta}{\theta}.\tag{2}$$ Now putting (1) and (2) together, we get $$\cos\theta\lt \frac{\sin\theta}{\theta}\lt \frac{1}{\cos\theta}.\tag{3}$$

Finally, let $\theta\to 0$. The left wall and the right wall in (3) both approach $1$, and poor $\frac{\sin\theta}{\theta}$ is therefore forced to approach $1$.

Answer 3

by the scheme we have that the area of triangle $OAD$ is $$A_1=\frac{\cos \theta\cdot \sin \theta}{2},$$

The area os sector $OAC$ is $$A_2=\frac{1}{2}R^2\theta=\frac{\theta}{2}, $$ because $R=1$.

The area of triangle $OCB$ is $$A_3=\frac{1\cdot \tan \theta}{2}=\frac{\tan \theta}{2}. $$ Now, note that $A_1\leq A_2\leq A_2$, i.e., $$\frac{\sin \theta\cdot \cos \theta}{2}\leq \frac{\theta}{2}\leq \frac{\tan \theta}{2}, $$ i.e., $$\sin \theta\cdot \cos \theta \leq \theta \leq \frac{\sin \theta}{\cos \theta}. $$ This answers the item (a).

For (b): dividing the inequality in (a) by $\sin \theta$ (which is positive because $0<\theta<\frac{\pi}{2}$,c.f. picture) we obtain $$\cos \theta\leq \frac{\theta}{\sin \theta} \leq \frac{1}{\cos \theta}$$ and "inverting", we obtain $$\cos \theta\leq \frac{\sin\theta}{\theta}\leq \frac{1}{\cos \theta} $$ By the theorem of "sandwich", passing the limit when $\theta\rightarrow 0^+$ we obtain that $$1\leq\lim_{\theta\rightarrow 0^+}\frac{\sin \theta}{\theta}\leq 1, $$ i.e., $$\lim_{\theta\rightarrow 0^+}\frac{\sin \theta}{\theta}= 1. $$