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Limit of (1-1)/h as h goes to 0?

  1. Nov 29, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the limit of (1-1)/h as h goes to 0?

    2. Relevant equations

    f'(x) = Lim(h→0) (f(x+h) - f(x))/h

    3. The attempt at a solution

    The answer is obviously 0, but I cannot figure out exactly how they got that answer.
     
  2. jcsd
  3. Nov 29, 2013 #2

    lurflurf

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    (1-1)/h=0
    for all nonzero h
    therefore
    $$\lim_{h\rightarrow 0}\frac{1-1}{h}=\lim_{h\rightarrow 0}0$$
     
  4. Nov 29, 2013 #3

    Student100

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    Gold Member

    I think I stole your name, or you stole mine. ;)

    Maybe it would also be helpful if you looked at this way:

    When thinking about limits, h is never actually 0. We're interested in what happens as h gets so close to zero we basically treat it as such. Since what you wrote is equivalent to saying [tex] \frac {0}{.000000....1} [/tex] you can conclude that this limit is 0.

    If on the other hand you had wrote:

    [tex] \lim_{h\rightarrow 0}\frac{(1+h)-1}{h}= 1 [/tex]

    Would be the correct interpretation of the limit.

    You'll learn more about limits when you discuss the epsilon/delta definition of a limit.
     
    Last edited: Nov 30, 2013
  5. Dec 10, 2013 #4
    Isn't the limit 0?
     
  6. Dec 10, 2013 #5

    lurflurf

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    The other student was comparing two different limits.

    $$\lim_{h \rightarrow 0} \frac{1-1}{h}=0 \\
    \lim_{h \rightarrow 0} \frac{1+h-1}{h}=1 \\
    \text{both special cases of}\\
    \lim_{h \rightarrow 0} \frac{1+a \, h-1}{h}=a$$
     
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