You can't divide by zero.
But, as you divide by numbers closer and closer to zero, the outputs blow up.
They get bigger and bigger, without bound.
Suppose a ladder is leaning against a wall, and you're working at the top of the ladder.
Someone (mischievous!) starts pulling the bottom of the ladder away at a constant rate.
The top of the ladder slides down the wall, with you on it.
You'll drop faster and faster as you near the ground!
(Interested? Search for the ‘falling ladder problem’. Don't try this!)
As introduced in the earlier section, Introduction to Asymptotes:
In particular, here is the definition of a vertical asymptote:
DEFINITION
vertical asymptote
A ‘vertical asymptote’ is a vertical line
that another curve gets arbitrarily close to
as $\,x\,$ approaches a finite number.
Specifically, the vertical line $\,x = c\,$ is a vertical asymptote for a function $\,f\,$ if and only if at least one of the following conditions is true:

The tangent function has infinitely many vertical asymptotes. For example, the vertical line $\,x = \frac{\pi}{2}\,$ is a vertical asymptote. as $\,x\rightarrow {\frac{\pi}{2}}^{+}\,,$ $\,\tan x\rightarrow \infty\,$ as $\,x\rightarrow {\frac{\pi}{2}}^{}\,,$ $\,\tan x\rightarrow \infty\,$ 
as $\,x\rightarrow 2^+\,,$ $\,y \rightarrow \infty$  as $\,x\rightarrow 2^+\,,$ $\,y \rightarrow \infty$  as $\,x\rightarrow 2^\,,$ $\,y \rightarrow \infty$  as $\,x\rightarrow 2^\,,$ $\,y \rightarrow \infty$ 
as $\,x\rightarrow 2^+\,,$ $\,y \rightarrow \infty$ as $\,x\rightarrow 2^\,,$ $\,y \rightarrow \infty$ 
as $\,x\,$ approaches $\,2\,$ from the right, $\,y\,$ goes to infinity 
as $\,x\,$ approaches $\,2\,$ from the right, $\,y\,$ goes to negative infinity 
as $\,x\,$ approaches $\,2\,$ from the left, $\,y\,$ goes to infinity 
as $\,x\,$ approaches $\,2\,$ from the left, $\,y\,$ goes to negative infinity 
How can a rational function (a ratio of polynomials) ‘blow up’ near a finite input?
When you're ‘trying’ to divide by zero!
Remember—dividing by a very small number gives a very big number.
Given a vertical asymptote, you usually want to know how the function behaves nearby.
Are the outputs going to infinity? To negative infinity?
Although you could certainly use a graphing calculator or WolframAlpha to see this behavior,
you should also be able to determine it algebraically, as shown next.
EXAMPLE:
The function $\displaystyle\,f(x) = \frac{3}{2x + 1}\,$ has a vertical asymptote
at $\,x = \frac 12\,.$
Why?
The denominator, $\,2x + 1\,,$ is zero when $\,x = \frac 12\,,$
and the numerator is (always) nonzero.
Here is the thought process for determining what the outputs from $\,f\,$ look like,
close to $\,\frac 12\,$:
Consider values of $\,x\,$ a bit less than $\,\frac 12\,$ (i.e., just to the left of $\,\frac 12\,$):
Observe that the notation $\,\frac{()}{(\text{small })}\,$ is being used to denote a (normalsized) negative number, divided by a small negative number—which produces a large positive number. Next, consider values of $\,x\,$ a bit more than $\,\frac 12\,$ (i.e., just to the right of $\,\frac 12\,$):
Observe that we did not compute actual values of the output near $\,x = \frac 12\,$! There's no need—it would be working too hard. All we need to know is:

On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
