# Math Help - Epsilon (Did I do this correctly)

1. ## Epsilon (Did I do this correctly)

I have a problem that says

f(x)= x^2 L=4 Xo= -2 e= 0.5

I have to find the value of delta.

Do I start by setting it up as

-.5 < x^2 -4 < .5??

sq root 3.5 < x < sq root 4.5

Then I take the larger absolute value of the 2....

sq root 4.5 - 2 = .12

Did I do this problem correctly?

2. You are correct in performing reversible operations all the way up to

$3.5 < x^2 < 4.5.$

Here, we must remember that $x$ could be positive or negative, and therefore that

$\sqrt{x^2}=|x|,$ not $x.$

Hence, we have

$\sqrt{3.5} < |x| < \sqrt{4.5}.$

3. Originally Posted by KarlosK
I have a problem that says

f(x)= x^2 L=4 Xo= -2 e= 0.5

I have to find the value of delta.

Do I start by setting it up as

-.5 < x^2 -4 < .5??

sq root 3.5 < x < sq root 4.5

Then I take the larger absolute value of the 2....

sq root 4.5 - 2 = .12

Did I do this problem correctly?
Take δ =1/2 => |x-(-2)| =|x+2|<1/2 => |x+2|<1 => |x|-2<1 => |x|<-1 =>|x|+2<1 => |x-2|<1 => |x+2||x-2|<1/2 => $|x^2-4|<\frac{1}{2}$

4. In the future, please state the problem completely. People who reponded guessed that you mean "Find $\delta> 0$ such that is $|x- 2|< \delta$ then $|x^2- 4|< \epsilon$" but they shouldn't have to guess!

And, of course, there exist an infinite number of answers to that question!

5. Originally Posted by HallsofIvy
In the future, please state the problem completely. People who reponded guessed that you mean "Find $\delta> 0$ such that is $|x- 2|< \delta$ then $|x^2- 4|< \epsilon$" but they shouldn't have to guess!

And, of course, there exist an infinite number of answers to that question!
I copied the problem exactly how it was worded in the book. I come on here to get math help, not to be talked down to. That would be like me saying please spell check your answer before posting it afterall I assume you meant responded by "reponded"

6. Originally Posted by KarlosK
f(x)= x^2 L=4 Xo= -2 e= 0.5
Start with bounding $|x+2|$ by 1. Why do that? Because $(x+2)$ is the other factor in $x^2-4$.
$\left| {x + 2} \right| < 1 \Rightarrow - 3 < x < - 1 \Rightarrow - 5 < x - 2 < - 3 \Rightarrow \left| {x - 2} \right| < 5$.
Find a delta that works $\delta = \frac{{0.5}}{5} = 0.1$
Note that if $|x+2|<\delta<1$ we get what we want.
$\left| {x^2 - 4} \right| = \left| {x - 2} \right|\left| {x + 2} \right| < (5)(0.1) = 0.5$.