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Epsilon Delta Proof of a Limit

  1. Oct 11, 2013 #1
    Hey there, I'm new to this forum. Today I thought I would brush up on my calculus.
    I would just like to know if my method is correct. Is there an easier way to prove this ?
    By the way, it's my first time using LaTeX, so bear with me.

    I am trying to prove the following :

    [tex]
    \lim_{x\rightarrow 10} {x^2} = 100
    [/tex]
    So, I must find a δ in which the following holds

    [tex]
    \forall\epsilon>0\ \exists\ \delta>0\ such\ that\ 0<|x - 10| < \delta \implies |x^2 - 100|< \epsilon
    [/tex]
    I observe the following
    [tex]|x-10| = |x + (-10)|[/tex] and by triangle inequality, [tex]|x|+|-10| > |x-10|[/tex] We will also note that [tex]|x| + |-10| = |x| + |10|\implies|x|+|10|>|x-10|[/tex]
    Now we can find a δ in terms of ε. By reverse triangle inequality,
    [tex]|x^2 - 100|\geq||x^2| - |100||\ and\ ||x^2| - |100||>|x^2| - |100|\ and \ |x^2| - |100| = |x|^2 - |10|^2[/tex][tex]\implies\epsilon > |x|^2 - |10|^2[/tex] Using this we can see that,[tex]\frac{\epsilon+|10|^2}{|x|}>|x|\ as\ well\ as \ \frac{\epsilon-|x|^2}{|10|^2}>|10| \ and\ since\ we\ know:\ |x|+|10| > |x-10|,[/tex]
    [tex]\implies \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|} > |x-10|[/tex]
    So take [tex]\delta = \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|}[/tex]

    Is this correct? I would be really grateful if I got some feedback!
    Can we also say the function is continuous [itex]\forall c\in\mathbb R[/itex] in the following way.
    [tex]\lim_{x\rightarrow c} {x^2} = c^2[/tex] just take [tex]\delta = \frac{\epsilon+|c|^2}{|x|} + \frac{\epsilon-|x|^2}{|c|}[/tex]
     
    Last edited: Oct 11, 2013
  2. jcsd
  3. Oct 11, 2013 #2

    Zondrina

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    Homework Helper

    That seems incredibly overcomplicated.

    ##|x^2-100| = |x-10||x+10| < δ|x+10|##

    Now by the triangle inequality:

    ##|x+10| = |x - 10 + 20| ≤ |x-10| + 20 < δ + 20##

    Now bound delta, what can you conclude?
     
  4. Oct 11, 2013 #3
    Can we just say [itex]\epsilon = \delta|x+10|\implies\frac{\epsilon}{|x+10|}=\delta[/itex]
     
  5. Oct 11, 2013 #4

    Zondrina

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    Bound delta by the simplest real number you can think of ##δ ≤ 1##.

    Now what can you conclude?

    ##δ|x+10| < δ(δ+20)##
     
  6. Oct 11, 2013 #5
    I see now ! [tex]take\ \delta ≤ 1\implies |x+10|≤ 1+2|10|[/tex] and so since we can say[tex]\epsilon = \delta |x+10| \implies \frac{\epsilon}{1+2|10|} = \delta[/tex]
     
    Last edited: Oct 11, 2013
  7. Oct 11, 2013 #6

    Zondrina

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    You could go as far as to say:

    ##δ(δ+20) ≤ 21δ## because ##δ ≤ 1##.

    Then you could conclude:
    ##21δ ≤ ε##
    ##δ ≤ \frac{ε}{21}##

    So ##δ = min\{1, \frac{ε}{21}\}##
     
  8. Oct 11, 2013 #7
    Thanks so your help I realized my answer was wrong after I posted it, Why is it we cannot use x to represent out epsilon ?
     
  9. Oct 11, 2013 #8

    Zondrina

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    ##x## has nothing to do with ##ε##. The objective is to get a ##δ(ε)##.
     
  10. Oct 11, 2013 #9
    I think I understand now, we are trying to find a bound on [itex]|x-c|[/itex] and it doesn't make sense to include [itex]x[/itex] in the value for [itex]\delta[/itex]
     
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