Evaluate the limit of $\arctan(\frac{1}{x - 2})$ where $x \to 2^+$. I'm having trouble understanding what my textbook is saying here. It seems to assign the variable $t$ out of nowhere, and I'm not sure why they do it.
Why assign $t$ to the same equation in $\arctan$?
My textbook says,

Evaluate the limit of $\arctan(\frac{1}{x - 2})$ where $x \to 2^+$.
If we let $t = 1 / (x - 2)$, we know that $t \to \infty$ as $x \to 2^+$. Therefore, by the formula $\lim_{x \to \infty} \tan^{-1}(x) = \frac{\pi}{2}$, we have
$$\lim_{x \to 2^+} \arctan(\frac{1}{x-2})= \lim_{t \to \infty} \arctan(t)=\frac{\pi}{2}$$

This is on page 132 of "Calculus: Early Transcendentals", 8th edition, by James Stewart.
 A: Forget about the $t$ for a second. What is going to happen to the fraction $\frac{1}{x-2}$ if $x$ is getting progressively closer to the number $2$ from the right? First of all, the $x$ is going to assume all possible values that are just a tad bigger than $2$: $2.1, 2.01, 2.001, 20001, 2.00001, 2.000001$, etc. When you subtract $2$ from something like $2.000001$, what do you get? You get a number that's going to be very, very small. And what happens when you divide 1 by a very, very small number? The result is going to be a huge number! This means that when $x$ approaches $2$ from the right, the fraction $\frac{1}{x-2}$ is going to positive infinity. And what does the inverse tangent function approach when its argument is going to positive infinity? The answer is $\frac{\pi}{2}$. That's why as $x\rightarrow2^+$, $\arctan\left(\frac{1}{x-2}\right)$ goes to $\frac{\pi}{2}$. This is absolutely equivalent to saying $\lim_{t\rightarrow\infty}\arctan{t}=\frac{\pi}{2}$ because $\frac{1}{x-2}$ in the arctangent function is going to infinity.
A: If you use
$\arctan(1/x)+\arctan(x)
=\pi/2$,
then
$\arctan(1/(x-2))
=\pi/2-\arctan(x-2)
$.
Since
$\lim_{x \to 0}\arctan(x)
=0
$,
$\lim_{x \to 2}\arctan(x-2)
=0
$,
so
$\lim_{x \to 2}\arctan(1/(x-2))
=\pi/2
$.
