Finding where a given piece-wise function. Is continuous.

Question

I'm a student studying math, and I'm going through some old exam problems and I have come across a set of questions that ask me to decide where a given piece-wise function is continuous. The function in question is.

$$f(x)= \begin{cases} \frac{x^4-1}{x^2-1} & \text{if }x \neq -1,1 \\ 2 & \text{if }x = 1 \\ -2 & \text{if }x =-1 \\ \end{cases} $$

At first glance it appears to be continuous everywhere except at $x=-1,1$ where the line breaks. I looked closely to weather the latter 2 conditions somehow patch the continuous line defined by the first condition but they didn't appear to.

I've only eyeballed it so if there is a rigours way that I should be using please let me know! It would be much appreciated!

Answer 1

The part of the function: $$f(x)=\dfrac{x^4-1}{x^2-1}$$ Can be simplified to: $$f(x)=x^2+1$$ Now it's just a simple matter if graphing the function. At $x=1$ we can see that the function is going to be continuous. At $x=-1,\dfrac{x^4-1}{x^2-1}\to 2$ but since the definition of the piece-wise function defines $f(-1)=-2$, the function cannot be continuous at $x=-1$

Answer 2

$$\displaystyle\lim_{x\to \pm 1}\frac{x^4-1}{x^2-1}=2$$ hence continuity fails at $x=-1$

Answer 3

For $x\neq \pm 1$, $x^2 \neq 1$ so we can simplify $\frac{x^4 - 1}{x^2-1} = x^2+1$. Draw the graph; $x^2+1$ at $x=-1$ is...? On the other hand the behaviour at $x=1$ is good.