How do I evaluate the following limit? 
I understand that I need to somehow use that $PR=AP=AQ$ as the point $A \to P$. But beyond that, I am unable to use that information to find $OB$.
This problem is from the textbook "Calculus with analytic geometry" by G. Simmons. The problem is the $27$th problem in section $12.2$ where the L'Hospitals rule is introduced.
 A: As the hint said, we have that $\triangle QAB$ and $\triangle PRB$ are similar.
If $\theta$ is expressed in radiants, we have that:
$$\bar {AP}=a \cdot \theta$$
And:
$$\bar{PR} =a\cdot \sin(\theta)$$
So, the simili-ratio is: $$r=\frac{\sin(\theta)}{\theta}$$
Now, we can write:
$$\frac{\bar{BR}}{\bar{AB}}=r\leftrightarrow \frac{x +a\cdot \cos(\theta)}{x+a}=r\leftrightarrow x(\theta-\sin(\theta))=a(\sin(\theta)-\theta\cos(\theta))\leftrightarrow x=\frac{a(\sin(\theta)-\theta\cos(\theta))}{\theta-\sin(\theta)}$$
Where $x=\bar{OB}$.
Now, we can take the limit as $\theta\to 0$, using Taylor series or L'Hospital rule as @insipidintegrator have done.
We have:
$$\lim_{\theta\to 0}\frac{a(\sin(\theta)-\theta\cos(\theta))}{\theta-\sin(\theta)}\,\,\sim\,\,\lim_{\theta\to 0}a\cdot\frac{\theta-\frac{1}{6}\theta^3-\theta+\frac{1}{2}\theta^3+o(\theta^3)}{\frac{1}{6}\theta^3+o(\theta^3)}=2a$$
In conclusion, when $\theta\to 0$ and $P\to A$, we have $\bar{OB}=2a$.
A: Note that $\theta\to 0,$ more precisely $0^+$, as $P$ approaches $A$ .

We’ll use coordinates here. Let the radius be $a$ so that $P\equiv (a\cos\theta,a\sin\theta),$ and $A\equiv (a,0)$. But the length of the arc $AP$ and hence that of segment $AQ$ is $a\theta$. Thus $Q\equiv (a, a\theta).$
Now the straight line through $P$ and $Q$ is
$$y-a\theta=(x-a)\left(\frac{a\theta-a\sin\theta}{a-a\cos\theta}\right)$$ which meets the X axis at   $B$ where $y=0$:$$x=a-a\theta\left(\frac {1-\cos\theta} {\theta-\sin\theta}\right)$$ Thus, $\displaystyle OB=\left| a-a\theta\left(\frac {1-\cos\theta} {\theta-\sin\theta}\right)\right|$ and we are asked to find $$L=\lim_{\theta\to 0^+} OB.$$ Thus, $$L=\left|\lim_{\theta\to 0^+} a-a\theta\left(\frac {1-\cos\theta} {\theta-\sin\theta}\right)\right|= \left|a-a\lim_{\theta\to 0^+}\theta\left(\frac {1-\cos\theta} {\theta-\sin\theta}\right)\right|$$ Now, we’ll use L’Hôpital’s Rule (I think you’re expected to use this because the rule has just been introduced) to evaluate the limit, which is in the $\displaystyle \frac 00$ indeterminate form: $$\lim_{\theta\to 0^+}\theta\left(\frac {1-\cos\theta} {\theta-\sin\theta}\right)= \lim_{\theta\to 0^+}\left(\frac {\theta-\theta\cos\theta} {\theta-\sin\theta} \right)$$$$=\lim_{\theta\to 0^+}\left(\frac {1-\cos\theta+\theta\sin\theta} {1-\cos\theta}\right)=\lim_{\theta\to 0^+}\left(1+\frac{\theta\sin\theta}{1-\cos\theta}\right)$$ Now use $\displaystyle \lim_{\theta\to 0^+}\frac{\sin\theta}{\theta}=1$ and $\displaystyle \lim_{\theta\to 0^+}\frac{1-\cos\theta}{\theta^2}=\frac 12$ to get the limit as 3. Thus, $$L=|a-3a|=|-2a|=2a.$$
