Question

Prove the limit as x approaches -2 of (x^2-1) = 3
This is the homework hints website but I'm not understanding all it's saying either. Thank you.

Answer 1

I'm not sure how to go about proving it other than by actually evaluating the limit at x = -2

lim ( (-2)^2 - 1) = 4 - 1 = 3
x -> -2

Answer 2

Thank you. This is some kind of proof where you find the delta along the x axis when given an epsilon along the y axis. Thank you for your time.

Answer 3

So, do you have the epsilon?

Answer 4

No, I believe they want me to solve for all epsilon > 0.

Answer 5

\[
|x-(-2)|<\delta \implies | (x^2-1) - 3|<\epsilon
\]

Answer 6

\[
|x+2|<\delta \implies | x^2-4 |<\epsilon
\]

Answer 7

\[

|x+2|<\delta \implies | (x-2)(x+2) |<\epsilon \\

|x+2|<\delta \implies | x-2||x+2 |<\epsilon
\]

Answer 8

Basically ideally we'd like to chose: \[
\delta \sim \frac{\epsilon}{|x-2|}
\]But we don't get to chose what \(x\) is. They do. We have to be safe for all cases.

Answer 9

Thank you for the detail.

Answer 10

To prevent \(\delta\) from getting to large, we'll say: \[
\delta \sim \min\{1,f(\epsilon)\}
\]

Answer 11

Now that we know \(\delta<1\) we can figure out what the range of \(|x-2|\) is.

Answer 12

\[
|x+2|<1\implies -1<x+2<1\implies -3<x<-1
\]Plugging those into \(|x-2|\): \[
|(-1)-2| = 3\\
|(-3)-2| = 5
\]This means: \[
|x-2|<5
\]

Answer 13

Likewise\[
|x-2|<5\implies \frac{\epsilon}{5}< \frac \epsilon {|x-2|}
\]

Answer 14

So our final answer to the challenge is: \[
\delta \sim f(\epsilon) =\min\left\{1,\frac \epsilon 5\right\}
\]

Answer 15

Finally, this completes the proof.