2
$\begingroup$

I'm having a hard time constructing a formal epsilon-delta proof for this given limit.

$$\lim_{x\to -2} \frac{x^2-4}{x+2} = -4$$

So I understood that I have to find what $δ$ should be for this implication to hold true: $$|x-(-2)|<δ \Rightarrow |\frac{x^2-4}{x+2}+4|<ε$$ $$|x+2|<δ \Rightarrow |\frac{x^2-4}{x+2}+4|<ε$$

I believe I have found what $δ$ must be in order for this implication to hold true, but I'm not sure if this is true.

This is what I have done to figure what $δ$ must be (this is my scratch work, not the proof): $$|\frac{x^2-4}{x+2}+4|<ε$$ $$|x-2|+4<ε$$ $$|x-2+4|<ε$$ $$|x+2|<ε$$ Therefore, I concluded that $δ = ε$.

Is this true? If this is true, how should my proof look like? I am really unsure of how to formally construct a proof using this information.

$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

In your scratch work, you should record the implications between each line.

The second line of your scratch work looks flawed.

If that second line were removed, then you could write implications as follows:

  • the first line is true $\iff$ the third line is true $\iff$ the fourth line is true

(These $\iff$ statements are all assuming that $x \ne -2$).

Next, set $\delta=\epsilon$.

Finally, assume $|x+2|<\delta=\epsilon$ and prove $|\frac{x^2-4}{x+2}+4|<ε$ by following the double implication arrows backwards from the last line of your scratch work to the first line.

$\endgroup$
3
  • $\begingroup$ I see! I had no idea this was an illegal step. So, how should a formal proof for this limit look like? Could you perhaps point me to the right direction on how to start it? $\endgroup$
    – vadimtimb
    Oct 12, 2019 at 19:54
  • 1
    $\begingroup$ In general you cannot replace $|a+b|$ with $|a|+b$. They are not equal in general, as you can easily tell by substituting some values. Although they are equal if $a,b$ are positive, one often uses absolute values in situations where the quantities are not positive. $\endgroup$
    – Lee Mosher
    Oct 12, 2019 at 19:59
  • $\begingroup$ As for the formal proof, I'll leave it to you to follow the instructions in my final paragraph. $\endgroup$
    – Lee Mosher
    Oct 12, 2019 at 20:00
2
$\begingroup$

Let $ \epsilon >0$ be given.

$(\star)$ $|\dfrac{x^2-4}{x+2}+4|=$

$|x-2+4|=|x+2|=|x-(-2)|$.

As suggested choose $\delta =\epsilon$.

Then

$|x-(-2)|<\delta$ implies

$(\star)$ $|\dfrac{x^2-4}{x+2}+4|= |x-(-2)|<\epsilon$.

$\endgroup$
4
  • $\begingroup$ So, how would I go about consturcting a formal proof for this limit? I am kind of lost as to how to formally write it up. I am new to proofs, and I don't really know how a formal proof for this kind of problem should look like. $\endgroup$
    – vadimtimb
    Oct 12, 2019 at 19:57
  • 1
    $\begingroup$ Harry.This is! a formal proof.1) $\epsilon >0$ given.2)Try to reduce $|f(x)+4| $to something involving $|x-(-2)|$ .Here both expressions turn out to be the same.3) So given $\epsilon$, choose $\delta=\epsilon$ to get from $|x-(-2)|<\delta$ to $ |x-(-2)|<\epsilon$.ok? $\endgroup$
    – Peter Szilas
    Oct 12, 2019 at 20:11
  • $\begingroup$ Oh, I apologize! I understand now. Thanks! $\endgroup$
    – vadimtimb
    Oct 12, 2019 at 20:18
  • $\begingroup$ Harry.A bit tricky to explain, do a few simple problems to get the idea. $\endgroup$
    – Peter Szilas
    Oct 12, 2019 at 20:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .