5

Can you actually cancel out the numerator and denominator?

$$f(x) = \frac{x^3-8}{x-2}$$

This function is not defined at $x=2$ so the domain of it is "all real numbers except 2". $f(x)$ can be rearranged into

$$\frac{(x^2+2x+4)(x-2)}{x-2}$$

When I cancel out the $x-2$, I get $x^2+2x+4$ which is defined at all real numbers. Let $g(x) = x^2+2x+4$. Are $f(x)$ and $g(x)$ different functions?

Am I correct to say $f(x)$ is not continuous at $2$, because I can't find $f(2)$?

I am curious about this because I saw this guy at http://www.youtube.com/watch?v=7tKq2NL0GJ4 You can scroll to 4:00 minute where he tries to find continuity of f(x) at x=2. I think that his method is incorrect because f(2) is undefined and you should answer that f(x) is not continuous x=2 immediately. Am I right?

Reinstate Monica -- notmaynard
  • 124
offchan
  • 478
  • 2
    MSE is quite serious already, wrong answers are commented to get edited and downvoted if that doesn't happen. – UserX Sep 18 '14 at 12:52
  • The question in bold is a good question. I think also you have received at least one good answer. – David K Sep 18 '14 at 13:11

2 Answers2

7

Good question. The functions $\frac{x^3-8}{x-2}$ and $x^2+2x+4$ are equal for all $x$ except at $x=2$, where the first function is undefined, and the second function is $12$.

However, if you are taking the limit of the first function as $x \to 2$, all you care about are the values of the function at $x$ near $2$; the value of $x$ at $2$ is irrelevant when taking the limit (even if it is undefined there like this case). So canceling is ok in this case.

angryavian
  • 89,882
  • I do not mean about finding its limit. I mean will f(x) stay the same after cancelling out the x-2 or not. In the video, she canceled it before taking limit. That changes the meaning of f(x). – offchan Sep 18 '14 at 13:01
  • Please change your answer according to my edited question if possible. Thank you. – offchan Sep 18 '14 at 13:05
  • I didn't ask about what is the limit of f(x). Please see the question again. – offchan Sep 18 '14 at 13:19
  • 1
    @off99555 My first paragraph answers your first question: they are different functions because they do not agree at $x=2$. Your justification for $f$ not being continuous at $x=2$ is correct. – angryavian Sep 18 '14 at 13:20
  • Thank you. I'm just not sure that you understand my question because I'm not a native English speaker and I can't type in English fluently. – offchan Sep 18 '14 at 13:27
4

You are right, $f$ and $g$ are different functions since they don't have the same domain.

And, yes, $f$ is not continuous at $2$ because it is not defined at $2$.

If we talk about limits then you can use the fact that if $f(x) = g(x)$ except at $x=a$, then $$ \lim_{x\to a} f(x) = \lim_{x\to a} g(x). $$ So now you note that for any $x\neq 2$, you indeed have $$ \frac{x^3 - 8}{x-2} = x^2 + 2x + 4. $$ And so $$ \lim_{x\to 2} \frac{x^3 - 8}{x-2} = \lim_{x\to 2} x^2 + 2x + 4 = \dots $$

Ilmari Karonen
  • 26,117
Thomas
  • 43,555
  • 1
    This is kind of good answer I'm looking for. Because I saw this guy at https://www.youtube.com/watch?v=7tKq2NL0GJ4 You can scroll to 4:00 minute where he tries to find continuity of f(x) at x=2. I think that his method is incorrect. Am I right? – offchan Sep 18 '14 at 13:20