Computing limits A circle of radius $1$ is centred at $(1, 0)$.  A chord of length $h$, $0 < h < 2$, intersects the circle at $(0, 0)$ and at the point $B$.  The point $A$ has coordinates $(0, h)$.  The line through $A$ and $B$ intersects the $x$-axis at a point with $x$-coordinate $P$.  Compute:  $ \lim_{h\to 0} P.$

So I'm torn on as $h$ approaches $0$, that the $P$ will go to $0$ or to $\infty$? I don't just want an answer but I want to gain the tools as to how to solve this problem. My calc teacher hasn's offered us the best tool to succeed. I'm willing to put in the work but don't know how to go about solving the issue.
Any help would be great.
 A: If $\theta$ is the angle between the line from the origin to $B$ and the $y$ axis.
$h = 2\sin \theta$
The angle at $P = \frac {\theta}{2}$
$P = h\cot (\frac {\theta}{2}) = 2\sin\theta \cot (\frac {\theta}{2}) = 4\cos^2 \frac {\theta}{2}$
$\lim_\limits{h\to 0} P = \lim_\limits{\theta \to 0} 4\cos^2\frac{\theta}{2}$
A: Here is a brute force approach.
The equation for the circle is $(x-1)^2+y^2=1$, so an equation for top half of the circle is $y=\sqrt{1-(x-1)^2}$, which simplifies to $y=\sqrt{2x-x^2}$.  If the coordinates of $B$ are $(x,y)=(x,\sqrt{2x-x^2})$, then $h^2=x^2+y^2=x^2+(2x-x^2)=2x$, so $h=\sqrt{2x}$, which means the coordinates of $A$ are $(0,\sqrt{2x})$.  The slope of the line connecting $A$ and $B$ is thus
$${\sqrt{2x-x^2}-\sqrt{2x}\over x-0}$$
But the slope of the line connecting $A$ and $P$ is 
$$0-\sqrt{2x}\over P-0$$
Since it's the same line, we can equate these and solve for $P$:
$${\sqrt{2x-x^2}-\sqrt{2x}\over x}={-\sqrt{2x}\over P}\implies P={x\sqrt{2x}\over\sqrt{2x}-\sqrt{2x-x^2}}$$
This now simplifies to
$$P={x\sqrt{2x}(\sqrt{2x}+\sqrt{2x-x^2})\over2x-(2x-x^2)}={2x^2+x\sqrt{4x^2-2x^3}\over x^2}=2+\sqrt{4-2x}$$
If you like, you can write this as $P=2+\sqrt{4-h^2}$.  The limit is now obvious:  as $h\to0$, $P\to2+\sqrt4=4$.
The relatively simple result suggests there ought to be a simpler way to obtain it, and this may be what the extra credit is hinting at, but I personally don't see it.  I hope someone else will.
