How Do You Solve a Delta/Epsilon Limit Proof Involving a Rational Function?

[iRaid]

Homework Statement

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

Homework Equations

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|&lt;\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|&lt;.05
\left|x+1-4\right|&lt;.05
\left|x-3\right|&lt;.05
-.05&lt;x-3&lt;.05
7.95&lt;x+5&lt;8.05
\delta=.05

Is that right :|PS: can someone tell me how to fix the limit in latex?

Thanks.

 

[Mark44]

Homework Statement

Find L=\lim_{x\rightarrow x_{0}} f(x). Then find a number \delta &gt; 0 such that for all x, 0&lt;\left|x-x_{0}\right|&lt;\delta \Rightarrow \left|f(x) - L\right|&lt;\epsilon

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

Homework Equations

The Attempt at a Solution

Found the limit first which = -4
\left|f(x) - L\right|&lt;\epsilon
\left|\frac{x^{2}+6x+5}{x+5} - 4\right|&lt;\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|&lt;.05
\left|x+1-4\right|&lt;.05
\left|x-3\right|&lt;.05
-.05&lt;x-3&lt;.05
7.95&lt;x+5&lt;8.05
\delta=.05

Is that right :|


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

Thanks.

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.

 

[iRaid]

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

Also, I only have 1 \ :|

 

[Mark44]

This is what you had:
[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?

 

[iRaid]

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

 

[Mark44]

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

Right. My previous post has been corrected.