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Delta epsilon proof

  1. Jan 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove [tex]lim_{x->\frac{1}{10}}\frac{1}{x}=10[/tex]

    2. Relevant equations
    |f(x)-L|<epsilon, |x-a|<delta

    3. The attempt at a solution
    I need to go from 1/x to x, so I applied an initial condition of delta<1/20
    [tex]
    \frac{-1}{20}<x-\frac{1}{10}<\frac{1}{20}[/tex]
    [tex]\frac{1}{10}<x<\frac{3}{10}[/tex]
    [tex]\frac{1}{x}<10
    [/tex]

    Moving on to the left side of the proof,
    [tex]
    |\frac{1}{x}-10|<\epsilon[/tex]
    [tex]-\epsilon<\frac{1}{x}-10<\epsilon[/tex]

    This is where I am stuck. If I use the fact that 1/x<10, I end up with -epsilon<10-10<epsilon, which isn't helpful.
     
  2. jcsd
  3. Jan 27, 2015 #2

    LCKurtz

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    I like that so far. Now look at your original problem$$
    \left| \frac 1 x - 10\right| = \frac{|1-10x|}{|x|}= \frac{|\frac 1 {10}- x|}{|\frac x{10}|}$$which you are trying to make small. Can you underestimate the denominator using your last line above?
     
    Last edited: Jan 27, 2015
  4. Jan 27, 2015 #3
    Okay, so I took |1/x-10|<epsilon and did

    [tex]\frac{1}{x}-\frac{10x}{x}<\epsilon[/tex]
    [tex]\frac{1-10x}{x}<\epsilon[/tex]
    [tex]\frac{-10(x-\frac{1}{10})}{x} <\epsilon[/tex]
    [tex]-10(x-\frac{1}{10})<\delta, \; \; \mbox{where} \; \; \delta=\frac{1}{20}[/tex]

    which leaves me with [tex]\frac{1}{200}<x-\frac{1}{10}<-\frac{1}{200}[/tex]
    ,Which is impossible. epsilon/200 seems to satisfy the requirement, though.

    Edit: Oops, I forgot to switch the inequality symbols. It should be

    [tex]\frac{1}{200}>x>-\frac{1}{200}[/tex]
     
    Last edited: Jan 27, 2015
  5. Jan 27, 2015 #4

    LCKurtz

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    I wasn't done editing and accidentally posted. Look at it now.
     
  6. Jan 27, 2015 #5
    I made a few errors in my post as well. I'm going to retype everything I've done, please let me know if I've reached a solution.

    [tex]|\frac{1}{x}-10|<\epsilon \; \; \mbox{when} \; \; |x-\frac{1}{10}|<\delta[/tex]

    Let δ=1/20. Then, expanding the left side, we get

    [tex] -\frac{1}{20}<x-\frac{1}{10}<\frac{1}{20} \; \rightarrow \; \frac{1}{10}<x<\frac{3}{10} \; \iff \; \frac{1}{10}<x \; \; \mbox {and} \; \; x<\frac{3}{10}[/tex]
    Inverting the left side, we get
    [tex]\frac{1}{x}<10[/tex]

    Moving on to the right side,

    [tex]
    \frac{1}{x}-\frac{10x}{x}<\epsilon \; \rightarrow \; \frac{1-10x}{x}<\epsilon \; \rightarrow \;
    \frac{-10(x-\frac{1}{10})}{x} <\epsilon[/tex]

    We assume that
    [tex]
    -10(x-\frac{1}{10})<\delta, \; \; \mbox{where} \; \; \delta=\frac{1}{20}[/tex]
    [tex]
    \frac{1}{200}>x-\frac{1}{10}>-\frac{1}{200} \; \iff \; x-\frac{1}{10}< \frac{1}{200}[/tex]
    Which implies that
    [tex]\delta=min \{ \frac{1}{20},\frac{1}{200} \} \; \rightarrow \; \delta=\frac{1}{200}[\tex]

    I'm assuming that I made a mistake on that last bit as 1/200 seems an odd bound.

    Edit: I'm not sure why that last bit of latex isn't working. It should look like http://latex.codecogs.com/gif.latex...ightarrow&space;\;&space;\delta=\frac{1}{200}
     
  7. Jan 27, 2015 #6

    LCKurtz

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    Your ##\delta## must depend on ##\epsilon##. I think if you are careful you will find, with your original assumptions that ##\delta =\frac \epsilon {200}## will work. I suggest you work with absolute values. Think about underestimating ##\frac {|x|}{10}## in the denominator as I suggested. It's only another step or two.
     
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