I'm trying to do an exercise but I honestly don't know what to do to solve it. I know this will be a very basic question, but I'm learning limits.

The exercise is:


$$\lim\limits_{x \to 2} \frac{f(x) - 6}{x^2-4} = 5$$

Find: $$\lim\limits_{x \to 2} f(x)$$

I tried separating the limit but I didn't get anything from it.

If you could please help me understand what to do, I'd be greatful (I need to understand, not just the answer).


1 Answer 1


Since$$\lim_{x\to2}\frac{f(x)-6}{x^2-4}=5,$$you have\begin{align}\lim_{x\to2}f(x)-6&=\lim_{x\to2}\left(\frac{f(x)-6}{x^2-4}(x^2-4)\right)\\&=\lim_{x\to2}\frac{f(x)-6}{x^2-4}\times\lim_{x\to2}(x^2-4)\\&=5\times0\\&=0.\end{align}So, $\lim_{x\to2}f(x)=6$.

  • $\begingroup$ Thank you! but what if I was asked a different limit? for instance $\lim\limits_{x \to 3} f(x)$ ? $\endgroup$
    – John Myers
    Apr 21, 2021 at 6:09
  • $\begingroup$ Then the answer would be $31$. $\endgroup$ Apr 21, 2021 at 6:11
  • $\begingroup$ but how would you do it? This example with limit x -> 3 is just a random number I typed, not an exercise I have. I just want to understand this throughly $\endgroup$
    – John Myers
    Apr 21, 2021 at 6:13
  • $\begingroup$ I wrote $31$ assuming that what you had in mind was that $\lim_{x\to3}\frac{f(x)-6}{x^2-4}=5$. Am I right? $\endgroup$ Apr 21, 2021 at 6:15
  • 2
    $\begingroup$ They can't (or, at least, they shouldn't). $\endgroup$ Apr 21, 2021 at 7:24

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