2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Paul Feb 13 at 12:02
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.