Help with epsilon and delta..

**Manizzle** · September 17th 2008, 06:08 PM

I am having a real tough time understanding problems dealing with epsilon and delta, so I am just picking problems out of my book and trying to do them myself but don't have solutions to them, so if someone could help me with this problem it would be great....

I am to prove the statement using the epsilon, delta definition of limit

limit as x --> 3 for (x^2 + x - 4) = 8

and sorry i don't have it in math format how you guys can type it..

**o_O** · September 17th 2008, 06:31 PM

Are you sure you wrote it correctly?

$\lim_{x \to {\color{red}2}}\left( x^2 + x - 4\right) = (2)^2 + 2 - 4 = 2 \quad \text{(not 8)}$

**Jhevon** · September 17th 2008, 06:33 PM

Originally Posted by Manizzle

I am having a real tough time understanding problems dealing with epsilon and delta, so I am just picking problems out of my book and trying to do them myself but don't have solutions to them, so if someone could help me with this problem it would be great....

I am to prove the statement using the epsilon, delta definition of limit

limit as x --> 3 for (x^2 + x - 4) = 8

and sorry i don't have it in math format how you guys can type it..

by the definition of "limit", you need to show that for all $\epsilon > 0$ , there exists $\delta > 0$ , such that $x \in \text{dom}(f)$ and $|x - 3| < \delta$ implies $|(x^2 + x - 4) - 8| < \epsilon$

so, set $|(x^2 + x - 4) - 8| < \epsilon$ , now you want to solve for $|x - 3|$ , then you can pick a reasonable $\delta$

**Manizzle** · September 17th 2008, 06:33 PM

sorry it was as x approaches 3..

**o_O** · September 17th 2008, 06:47 PM

I think you mean as x approaches 3 right? It doesn't work if x approached 2.

$\lim_{x \to {\color{red}3}}\left( x^2 + x - 4\right) = 8$

Going by definition:
$\forall \epsilon > 0, \ \exists \delta > 0$ such that whenever $0 < | x - 3 | < \delta$ it follows that $\left|x^2 + x - 4 - 8\right| < \epsilon$ .

Looking at that last expression:
$\begin{array}{rcl} \left|x^2 + x - 12\right| & < & \epsilon \\ \left|x - 3\right| \left|x + 4 \right|& < & \epsilon \qquad {\color{red}(1)} \end{array}$

Since we're dealing with values of x near 3, we can assume that the distance from these values to 3 is less than 1, i.e. $|x - 3| < 1$

How does this affect the value of $|x + 4|$ ? Well:
$|x - 3| < 1 \ \iff \ -1 < x - 3 < 1 \ \iff \ -1 + 7 \ < \ x + 4 \ < \ 1 + 7$ ${\color{white}.} \ \iff \ 6 < x + 4 < 8 \ \Rightarrow \ | x + 4 | < 8$

Try to use this fact about $|x + 4|$ in ${\color{red}(1)}$

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