Identify the value of δ( accur. to 2 dec. places) that corresponds to ε=0.01?

[solve]

Homework Statement

Identify the value of δ( accur. to 2 dec. places) that corresponds to ε=0.01, given lim x->-1 (x^2+3)=4, according to the definition of limits.

Homework Equations

|f(x)- L| < ε

0<|x-a| < δ

The Attempt at a Solution

|(x^2+3)- 4| <0.01

|x^2-1| <0.01

|x-1||x+1| <0.01

This where I have no idea what to with |x-1|. Do I divide 0.01 by |x-1|? Any hints, please?

Thanks.

 

[HallsofIvy]

You don't want to divide by |x- 1|, you want to divide by a number which means that you need to replace |x- 1| by a number that keeps the inequality true.

You only need to prove this for x "close" to -1 so start by assuming that |x-(-1)|= |x+ 1|< 1 (1 is chosen just as a convenient value- any positive number would work). That is, that -1< x+1< 1. Subtracting 2 from each part, -3< x- 1< -1 which is the same as saying that |x- 1|> 1. We want ">" here because now we can say that if we replace |x- 1| by 1 in |x- 1||x+ 1| we have replace it by something larger: |x-1||x+1|< |x+ 1|. If we take |x+1|< .001 then it is certainly true that |x-1||x+1|< |x+1|< .001.

 

[solve]

Hi, HallsOfIvy. I appreciate your answer. Just never got around to answering to this thread. Wanted to acknowledge I read your answer and probably will have some questions once I finally have time to look at it a bit closer. Thank You, HallsOfIvy.