# Homework Help: Delta/Epsilon Limit Proof

1. Oct 12, 2011

### iRaid

1. The problem statement, all variables and given/known data
Find $L=\lim_{x\rightarrow\x_{0}} f(x)$. Then find a number $\delta > 0$ such that for all x, $0<\left|x-x_{0}\right|<\delta$ $\Rightarrow$ $\left|f(x) - L\right|<\epsilon$

Problem:
$f(x)=\frac{x^{2}+6x+5}{x+5}$, $x_{0}=-5$, $\epsilon=0.5$
2. Relevant equations

3. The attempt at a solution
Found the limit first which = -4
$\left|f(x) - L\right|<\epsilon$
$\left|\frac{x^{2}+6x+5}{x+5} - 4\right|<\epsilon$ <--- Problem here not sure... My teacher seems to sometimes keep the negative limit or sometimes he'll make it positive
$\left|\frac{(x+5)(x+1)}{x+5} - 4\right|<.05$
$\left|x+1-4\right|<.05$
$\left|x-3\right|<.05$
$-.05<x-3<.05$
$7.95<x+5<8.05$
$\delta=.05$

Is that right :|

PS: can someone tell me how to fix the limit in latex?

Thanks.

Last edited: Oct 12, 2011
2. Oct 12, 2011

### Staff: Mentor

Removed an extra \.

First of all, your function is defined everywhere except at x = -5. For any other value of x, f(x) = x + 1. For this reason, $\lim_{x \to 5} f(x) = 6$, not -4.

3. Oct 12, 2011

### iRaid

oops, x0 is supposed to equal -5 not 5.

Also, I only have 1 \ :|

4. Oct 12, 2011

### Staff: Mentor

[noparse] $L=\lim_{x\rightarrow\x_{0}} f(x)$ [/noparse]

The \ after rightarrow and before x_{0} was causing the problem.

Your value of δ is not right.

Let's sum up, with errors corrected.
$$\lim_{x \to -5}f(x) = -4$$

For x $\neq$ -5, f(x) = [(x + 5)(x + 1)]/(x + 5)] = x + 1

You want to find δ so that if |x - (-5)| < δ, then |f(x) - (-4)| < 0.5.

Can you take it from here?

Last edited: Oct 13, 2011
5. Oct 13, 2011

### iRaid

-5+1=-4 not 6 tho >.<

6. Oct 13, 2011

### Staff: Mentor

Right. My previous post has been corrected.