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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.

dmtri
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Ehab
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    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 '19 at 12:02

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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: $$\hat{y}(x)=\begin{cases}2&\text{ if }x=1\\\frac{x^{2}-1}{x-1}&\text{ else}\end{cases}.$$

More on this can be found at: https://en.wikipedia.org/wiki/Classification_of_discontinuities.

Floris Claassens
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