Limits of Rational Functions

I may end up writing more than 2 blog posts per week. I’m not trying to be an overachiever (Ok maybe I am). I think that by writing about what we learned in the last class, I can help solidify it in my mind and perhaps my viewpoint could help a classmate. So here’s an update from last class.

#1. I got my quiz back and got a 10 out of 10! Woohoo!

#2. We learned about the Limits of Rational Functions and here’s what we learned:

Properties of Rational Functions:

Rational Functions are continuous everywhere that they are defined.
They can have Vertical Asymptote discontinuities OR
They can have Point Discontinuities
Rational Functions typically look something like this $f(x) = \dfrac{x-2}{x^2-2x}$

What’s important to remember, is that rational functions are continuous everywhere EXCEPT where the denominator is equal to zero. Aka if you solve for x on the above equation, and you get $\frac{2}{0}$ you have a discontinuity.

But what does that mean for limits? Well, let’s look at the two types of discontinuities we are dealing with for a rational function.

Vertical Asymptote: A vertical asymptote occurs when a graph has two lines that never cross a certain point of x. It looks a lot like this:

Thank you http://www.analyzemath.com for the photo

Please note that this discontinuity of a function means the function has no existing limit.

Point Discontinuity: This is where a point on the function is not passed through by the line on the graph. For example:

Thanks to Wikipedia for the image

Note that at $x_0$ there is an unfilled circle, that is the point discontinuity. It is important to know that despite this discontinuity, a limit still exists for this equation, because limits do not care about the exact point, but only leading up to the point.

So what we’ve reached thus far is that when dealing with a vertical asymptote, a limit will not exist, but when dealing with a point discontinuity a limit will exist.

How to find the limit/discontinuities of an ungraphed equation:

Let’s take the equation $f(x)=\dfrac{x-2}{x^2-2x}$

*In order to find the discontinuities, set the denominator to zero.

$x^2-2x=0$ $x(x-2)=0$ $x=0 or x=2$ the discontinuities are at x=0 and x=2

you can also simplify the function as an algebraic expression to find g(x) like so:

$f(x)=\dfrac{x-2}{x^2-2x}$ = $g(x)=\dfrac{x-2}{x(x-2)}=\frac{1}{x}$ so now $g(x)=\frac{1}{x}$ f(x)=g(x) everywhere they are DEFINED but they are not the same function.

Once you’ve simplified the function, you can use the simpler function to find the limit of the more complicated function.

Typically, if you do this, and can cancel out a factor, it’s a point discontinuity, but otherwise it is a vertical asymptote and no limit will exist. If you substitute your c into a rational function and you get $\frac{n}{0}$ where n is any non zero number, then there is a vertical asymptote at x=c and the limit of f(x) at x→c does not exist.

If you get $\frac{0}{0}$ factor it out into a g(x) and you should be able to find a limit.

If this seems like a lot of gibberish, the best way I can show you is by doing 2 problems and explaining it that way. Here we go:

$\lim_{x\to 3} \frac{x^2-8x+15}{x^2-7x+12}= \frac{0}{0}$ which means it’s a point discontinuity and we can solve it using a factored version of the equation: