Proving $\lim_{x\to1} f(x)=f(1)$ where $f(x)=\frac{3x+1}{x^2-4x-5}$

Question

I got stuck on the following problem

I have to prove that $f(x)=\frac{3x+1}{x^2-4x-5}$ is continuous at the point $x_0=1$.

What I've got:

First I consider that I don't want the denominator to be zero so I fix $|x-1|<1$. The denominator is always negative with that condition.

Let $\epsilon$ be a positive real number. I get: $$ |f(x)-f(1)|=\frac{|x+3||x-1|}{|x^2-4x-5|}<\epsilon. $$

I know that $|x+3| < 5$ but I don't know how to deal with the denominator so that I find sufficient conditions so that I fix a real $\delta(\epsilon)$.

Thanks in advance!

Answer 1

You are on the right track. We may assume that $1/2<x<3/2$ (so $0<\delta<1/2$), then $$|f(x)-f(1)|=\frac{|x+3||x-1|}{|(x+1)(x-5)|}\leq \frac{(3/2+3)}{(1+1/2) (5-3/2)}\cdot|x-1|.$$ Can you take it from here?

Answer 2

$$f(x)$$ is a continuous function, since it is a quotient of two continuously functions, namely $$x+1$$ and $$x^2-4x-5$$