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Why is the graph of $y=(x^2+x-6)/(x-2)$ the same as the graph of $y=x+3$, and a continuous function?

I understand that this is because $$ \frac{x^2+x-6}{x-2} = \frac{(x+3)(x-2)}{x-2} = x+3. $$ But my problem with this is that in $(x-2)$, when $x=2$ the denominator will be $0$. So shouldn't there be a break here, because if $y=(x^2+x-6)/(x-2)$, then $y$ should then be undefined? So why can $y=(x^2+x-6)/(x-2)$ still be considered continuous and equal to $y=x+3$?

Please try to explain this simply. I'm still just beginning Khan Academy calculus. Thank you.

Bilbottom
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  • $\lim_{x\to 2}\frac{x^2+x-6}{x-2}=5$ – A. Goodier Aug 22 '18 at 10:40
  • A graph says nothing. As you know, the two graphs should be the same everywhere $x\ne 2$; although the graphs are different at $x=2$, you won’t be able to see it because it is just a point. Well, this is not a mathematical explanation, but you should remember that plotting functions only gives you intuition but not proofs. – Lee Aug 22 '18 at 10:41

5 Answers5

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Their graphs are not the same, nor is $f(x) = (x^2+x-6)/(x-2)$ continuous at $x=2$.

Indeed, as you say $f(x) = (x^2+x-6)/(x-2)$ is undefined at $x=2$ and therefore has a sort of a "puncture" at $x = 2$.

In particular, $f$ cannot be continuous at $x=2$, since the definition of continuity at a point requires the function to be defined at that point.

MSDG
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You're right that $$ y = \frac{x^2+x-6}{x-2} $$ has a discontinuity at $x = 2$, so as it is, it isn't truly continuous. However, you've shown that this is equal to $y = x + 3$ for every $x \neq 2$, and by looking at the graph of $$ y = \frac{x^2+x-6}{x-2} $$ it makes sense that at $x = 2$, we should have $y = 5$ (when you come to learn about limits, this will make sense and can be proved formally). So, what we should actually do is write $$ y = \begin{cases} \frac{x^2+x-6}{x-2}, & x \neq 2,\\ 5, & x = 2.\\ \end{cases} $$ Appending this value makes it continuous everywhere, and makes it identically equal to $y = x + 3$.

Bilbottom
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Function $x\mapsto\frac{x^2+x-6}{x-2}=\frac{(x+3)(x-2)}{x-2}$ is defined on $\mathbb R-\{2\}$.

Function $x\mapsto x+3$ is defined on $\mathbb R$.

So the functions are not the same, but the second function restricted to set $\mathbb R-\{2\}$ is the same as the first.

The first function is continuous at any element of $\mathbb R-\{2\}$ hence is a continuous function.

We cannot say that it is continuous at $2$ because it is not defined there, but what we can say is that it can be extended to a continuous function on $\mathbb R$ (which is the second function).

For a related question that I posed myself see here.

drhab
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To elaborate on rbird's answer, if you have a $0/0$ singularity, you can use L'Hopital's rule to find the solution:

$$ \lim_{x\rightarrow c} \frac{f(x)}{g(x)}= \lim_{x\rightarrow c} \frac{f'(x)}{g'(x)} $$

and so

$$ \lim_{x\rightarrow2} \frac{x^2 +x-6}{x-2} = \lim_{x\rightarrow2}\frac{2x+1}{1} $$

which equals 5 QED

user68150
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An answer for you after learning complex analysis in the future

Let $f(z)=\frac{z^2-z+6}{z-2}$ and $g(z)=z-3$.

$f(z)$ has a domain of $\mathbb C\setminus\{2\}$.

$g(z)$ has a domain of $\mathbb C$.

Since $f(z)-g(z)=0$ for every $z\in\mathbb C\setminus\{2\}$, by the principle of analytic continuation, $g$ is the unique analytic continuation of $f$. This explains why the graphs look the same.

Lee
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