Epsilon (Did I do this correctly)

**KarlosK** · February 11th 2010, 08:47 AM

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?

**Scott H** · February 11th 2010, 05:33 PM

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

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

**xalk** · February 12th 2010, 01:07 AM

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 => $|x^2-4|<\frac{1}{2}$

**HallsofIvy** · February 12th 2010, 02:14 AM

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!

**KarlosK** · February 12th 2010, 08:11 AM

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"

**Plato** · February 12th 2010, 08:48 AM

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.
$\left| {x^2 - 4} \right| = \left| {x - 2} \right|\left| {x + 2} \right| < (5)(0.1) = 0.5$ .

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