# Difference between normal functions and discontinuous functions?

For the function, $$y=\frac{x^2-1}{x-1}$$ The denominator cannot be zero. So

$$\lim_{x\to1}\frac{x^2-1}{x-1}=\lim_{x\to1}(x+1)=2$$

"$$y=\frac{x^2-1}{x-1}$$ is discontinuous at $$x=1$$ since $$y$$ is undefined at that point. This leaves a gap in the curve. The limit tells us that $$y\to2$$ as $$x\to1$$, so the gap is at $$(1,2)$$ ."

This is a bit from my maths textbook (Maths In Focus) about discontinuous functions.

This is cool and good. However, what I'm having a bit of trouble with is understanding how can $$y=\frac{x^2-1}{x-1}$$ be equal to $$y=x+1$$ when they generate different graphs.

The graph $$y=\frac{x^2-1}{x-1}$$ is discontinuous while $$y=x+1$$ is continuous. What I don't understand is why does the graph of $$y=x+1$$ change when it is multiplied by $$\frac{x-1}{x-1}$$, which is essentially multiplication by one. How can you change a value/graph when all you do is multiply by one?

I have searched over the internet and there isn't a single article/video explaining this specifically, which probably means I'm misunderstanding something or overlooking something fundamental. Any clarification on what exactly is going on would be deeply appreciated.

• You are not multiplying by 1 when x = 1 though. Your function is $\frac{x^2-1}{x-1} = x+1$ when x is not 1 and undefined when x is 1. This is indistinguishable from $x+1$ as a picture but technically it has a diferent domain to your original function, hence is different. – Paul Feb 13 at 12:02

The problem is that $$\frac{x-1}{x-1}$$ is essentially 1, except for $$x=1$$ in which case the graph is undefined. This is called removable discontinuity, which means you can create a continuous function defined on the entirety of $$\mathbb{R}$$ by redefining $$y$$ in one point: $$\hat{y}(x)=\begin{cases}2&\text{ if }x=1\\\frac{x^{2}-1}{x-1}&\text{ else}\end{cases}.$$