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Delta/Epsilon Limit Proof

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
    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]
    2. Relevant equations



    3. 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: Oct 12, 2011
  2. jcsd
  3. Oct 12, 2011 #2

    Mark44

    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, [itex]\lim_{x \to 5} f(x) = 6[/itex], not -4.
     
  4. Oct 12, 2011 #3
    oops, x0 is supposed to equal -5 not 5.

    Also, I only have 1 \ :|
     
  5. Oct 12, 2011 #4

    Mark44

    Staff: Mentor

    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: Oct 13, 2011
  6. Oct 13, 2011 #5
    -5+1=-4 not 6 tho >.<
     
  7. Oct 13, 2011 #6

    Mark44

    Staff: Mentor

    Right. My previous post has been corrected.
     
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