Delta Epsilon Proofs involving rational functions

Question

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.

Answer 1

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.

Answer 2

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

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