# Find Discontinuity And Classifying Them

• May 13th 2012, 06:16 AM
Bashyboy
Find Discontinuity And Classifying Them
The problem I am looking at is $f(x) = \frac{x - 3}{x^2 - 9}$

I found that f was not defined at $x = \pm 3$

I have -3 as a non-removable discontinuity, and 3 as a removable discontinuity, approaching positive infinity from either side. Yet, when I graph the function, it is unequivocally approaching a finite value. Why did limiting process give me positive infinity?
• May 13th 2012, 06:26 AM
Prove It
Re: Find Discontinuity And Classifying Them
Quote:

Originally Posted by Bashyboy
The problem I am looking at is $f(x) = \frac{x - 3}{x^2 - 9}$

I found that f was not defined at $x = \pm 3$

I have -3 as a non-removable discontinuity, and 3 as a removable discontinuity, approaching positive infinity from either side. Yet, when I graph the function, it is unequivocally approaching a finite value. Why did limiting process give me positive infinity?

Notice that if \displaystyle \begin{align*} x \neq 3 \end{align*}, we have

\displaystyle \begin{align*} \frac{x - 3}{x^2 - 9} &\equiv \frac{x - 3}{(x - 3)(x + 3)} \\ &\equiv \frac{1}{x + 3} \end{align*}

The left hand limit as \displaystyle \begin{align*} x \to -3 \end{align*} is \displaystyle \begin{align*} -\infty \end{align*} while the right hand limit as \displaystyle \begin{align*} x \to -3 \end{align*} is \displaystyle \begin{align*} +\infty \end{align*}. Since the left and right hand limits are not equal, this is a jump discontinuity.
• May 13th 2012, 06:36 AM
Bashyboy
Re: Find Discontinuity And Classifying Them
Oh, so you can cancel out factors; I thought I remembered my teacher saying we couldn't, or maybe that was just in the original function.
• May 13th 2012, 06:51 AM
Prove It
Re: Find Discontinuity And Classifying Them
Quote:

Originally Posted by Bashyboy
Oh, so you can cancel out factors; I thought I remembered my teacher saying we couldn't, or maybe that was just in the original function.

You can cancel out factors as long as you realise that there is an extra discontinuity in your original function.