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Delta/Epsilon Limit Proof

  • Thread starter iRaid
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  • #1
559
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Homework Statement


Find [itex]L=\lim_{x\rightarrow\x_{0}} f(x)[/itex]. Then find a number [itex]\delta > 0[/itex] such that for all x, [itex]0<\left|x-x_{0}\right|<\delta[/itex] [itex]\Rightarrow[/itex] [itex]\left|f(x) - L\right|<\epsilon[/itex]

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

Homework Equations





The Attempt at a Solution


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

Is that right :|


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

Thanks.
 
Last edited:

Answers and Replies

  • #2
33,641
5,306

Homework Statement


Find [itex]L=\lim_{x\rightarrow x_{0}} f(x)[/itex]. Then find a number [itex]\delta > 0[/itex] such that for all x, [itex]0<\left|x-x_{0}\right|<\delta[/itex] [itex]\Rightarrow[/itex] [itex]\left|f(x) - L\right|<\epsilon[/itex]

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

Homework Equations





The Attempt at a Solution


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

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, [itex]\lim_{x \to 5} f(x) = 6[/itex], not -4.
 
  • #3
559
8
oops, x0 is supposed to equal -5 not 5.

Also, I only have 1 \ :|
 
  • #4
33,641
5,306
This is what you had:
[noparse] [itex]L=\lim_{x\rightarrow\x_{0}} f(x)[/itex] [/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.
[tex]\lim_{x \to -5}f(x) = -4[/tex]

For x [itex]\neq[/itex] -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:
  • #5
559
8
-5+1=-4 not 6 tho >.<
 
  • #6
33,641
5,306
-5+1=-4 not 6 tho >.<
Right. My previous post has been corrected.
 

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