# Decide where $f(x) = \frac{x +3}{x^2-1}$ for $x \neq -1,1$. Is continuous.

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 function is continuous . At first glance it appears the example above is continuous everywhere as $$x = -1,1$$. Is it as simple enough just to say, or am I missing something, like a rigours theorem of some sort?

Any help would be much appreciated.

• Your first glance should be correct. – Mathsisfun May 22 '20 at 3:57

## 1 Answer

The sum and product of continuous function is a continuous function, so it's enough to say that it will be continuous except for $$x=-1, x=1$$ since the function is undefined at those points

• So you say $\frac{x +3}{x^2-1}$ is continuous because ${x +3}$ and $\frac{1}{x^2-1}$ are continuous. Then the next question is, why is $\frac{1}{x^2-1}$ continuous? – miracle173 May 22 '20 at 5:27
• $f(x)=x^2-1$ & $g(x)=1$ is continuous, so $g/f$ is continuous where $f \ne 0$. – Tamas Kanti Garai May 22 '20 at 6:14