Delta-Epsilon Proofs and Bounds

Question

I'm taking a Calculus I course online, which has been going pretty well considering that I haven't done much math since I graduated high school 5 years ago. However, I recently ran into a good bit of trouble with the chapter on delta-epsilon proofs. Specifically, my question is about non-linear equations and bounds. Here's the example that I'm working off of:

Prove that $\lim\limits_{x \to -1} {x^2-2x-2} = 1$

So we can apply part of our delta-epsilon definition:

|$(x^2-2x-2)-1$| < $\epsilon$

Simplify:

|$x^2-2x-3$| < $\epsilon$

Factor:

|$(x-3)(x+1)$| < $\epsilon$

Here is where I get confused, because the textbook has no information about this particular problem! I tried to find some help online, and I have found some videos, but I am still not clear on what exactly I need to do next and why. I have read some web pages that talk about "bounding" the variables.

1) Why do we need to put a bound on these variables? 2) How do we bound them?

Answer 1

Intuition: $x$ being close to $-1$ will give you a bound on $x+1=x-(-1)$. But this in fact also gives you a bound on $x-3$ in the sense that $x-3$ is close to $-1-3=-4$. Say, if $x$ is within $1$ unit from $-1$, then $x-3$ is between $-5$ and $-3$.

Let $\delta=\min\{1,\epsilon/5\}$. Then if $|x-(-1)|<\delta$, then $$ |x+1|<\delta\leq\epsilon/5,\quad|x-3|=|x+1-4|\leq|x+1|+|-4|< 1+4=5. $$ Together, these imply $$ |x-(-1)|<\delta\implies|(x-3)(x+1)|=|x-3|\times|x+1|<5\times(\epsilon/5)=\epsilon. $$

Answer 2

A change in $x$ implies a change in $f(x)$.

$|x-x_{0}| \lt \delta \Rightarrow |f(x)-L| \lt \epsilon$

Know $|x--1| = |x+1| \lt \delta$

$f(x) = x^{2}-2x-2$

$L = 1$

$f(x)-L = x^{2}-2x-2-1 = (x-3)(x+1) \lt (x-3) \delta$ $\lt \epsilon$

The $(x-3)$ is almost a constant multiplier times by the decreasing variable $\delta$.