Compute Limit Question with tangent $$\lim_{x \to 0^+} \tan^{-1}\ \left(\frac{1}{x}\right)\
  $$
I am not sure how to solve this limit either, it says i should first see what 1/x is approaching but im confused as how to do that, and how to solve it. How would i rewrite this or what rules would i use?
 A: This problem can be rewritten by changing  $\lim_{x \to 0^+}$ to $\lim_{x \to \infty}$   and $\frac 1x$  to $(x)$.
A: As a positive number $x$ approaches 0, $1/x$ tends to infinity (since the numerator is a positive constant and the positive denominator is decreasing). Thus, $tan^{-1} (1/x)$ tends to $\pi/2$. For the latter, graph $y=tan^{-1}(x)$ and see that the line $y=\pi/2$ is the horizontal asymptote which $y$ approaches as the input of the function tends to infinity. 
A: As $x\to 0^+$, $x$ is always positive, but it’s getting smaller and smaller. Consequently, $\frac1x$ is also positive, but it’s getting bigger and bigger, and $\frac1x\to\infty$. It would probably help you to draw a graph of $y=\frac1x$, or display it on a graphing calculator: you’d see that as $x$ moves leftwards towards $0$, $\frac1x$ shoots rapidly upwards. There’s a picture here.
Now take it a step further. Let $y=\frac1x$, and ask yourself what happens to $\tan^{-1}y$ as $y\to\infty$. Recall that $\tan^{-1}y$ is the angle (in the first or fourth quadrant) whose tangent is $y$; what kinds of angles have very large tangents? Let’s see: $\tan\frac{\pi}6=\frac12,\tan\frac{\pi}4=1$, and $\tan\frac{\pi}3=\sqrt3\approx1.732$, to take a few familiar angles, so the tangent seems to be getting bigger as the angle increases. In fact the tangent an angle in the first quadrant is simply the slope of the line through the origin that makes that angle with the positive $x$-axis. As that slope gets bigger and bigger, what’s happening to the angle between the line and the $x$-axis? It’s approaching a right angle, or $\frac{\pi}2$.
