# Help with epsilon and delta..

• Sep 17th 2008, 07:08 PM
Manizzle
Help with epsilon and delta..
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..
• Sep 17th 2008, 07:31 PM
o_O
Are you sure you wrote it correctly?

$\lim_{x \to {\color{red}2}}\left( x^2 + x - 4\right) = (2)^2 + 2 - 4 = 2 \quad \text{(not 8)}$
• Sep 17th 2008, 07:33 PM
Jhevon
Quote:

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 $\delta$
• Sep 17th 2008, 07:33 PM
Manizzle
sorry it was as x approaches 3..
• Sep 17th 2008, 07:47 PM
o_O
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 ${\color{red}(1)}$