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Homework Help: Derivative help

  1. Jan 31, 2006 #1
    If https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image1.png [Broken] , find [PLAIN]https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image2.png [Broken] [Broken]using the definition of derivative. [PLAIN]https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image2.png [Broken] [Broken] is the limit as https://webwork2.math.ohio-state.edu/courses/math151wi06sm/tmp/png/Hmwk3/391484/kopko.4-prob17image4.png [Broken] of the expression (answer)
    I need help with finding the expression.

    The value of the limit is 0.

    Any help would be great.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 31, 2006 #2


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    What have you done?
  4. Jan 31, 2006 #3


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    f'(x0) is defined to be
    [tex]f'(x_0) = \lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}[/tex]
    Apply that to your function:
    [tex]f'(9) = \lim_{h \rightarrow 0} \frac{f(9 + h) - f(9)}{h}[/tex]. Now, how can you do this limit?
  5. Jan 31, 2006 #4
    I am just confused with the 1/2^.5 part.

    I am used to seeing something like f(x)= 1/x....so that you make it 1/(9+h).
  6. Jan 31, 2006 #5


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    [tex]f(x) := \frac{1}{\sqrt{2}}[/tex] means that this function is a constant and does not depend on what x you choose.
    It will return the same value (i.e, 1 / (2.5)), no matter what x is. For example:
    If [tex]g(x) := \frac{1}{\sqrt{3}}[/tex], then:
    [tex]g(0) = \frac{1}{\sqrt{3}}[/tex],
    [tex]g(7) = \frac{1}{\sqrt{3}}[/tex],
    [tex]g(9) = \frac{1}{\sqrt{3}}[/tex],...
    Do you understand? Can you solve the problem now?
  7. Jan 31, 2006 #6
    And if the limit definition doesn't click, look at it graphically. A constant function is just a horizontal line, which obviously has a slope of 0.
  8. Jan 31, 2006 #7
    I have looked at it graphically, but I having a hard time understanding how to make it into an expression...lim h->0 something.

    thanks for all of the input, but I am still confused.
  9. Jan 31, 2006 #8
    Ok. VietDao29 showed you the limit definition of a derivative. It's the following...

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

    Your function is [tex]\frac{1}{\sqrt{2}}[/tex], so I'll start you out.

    [tex]f'(9)=\lim_{h\rightarrow{0}} \frac{f(9+h)-f(9)}{h}[/tex]

    This is still from VietDao29's post.

    What is f(9)? What is f(0)? What is f of anything? If you can answer that, you can do your limit.
  10. Jan 31, 2006 #9


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    Another hint is that:
    [tex]\frac{f(9 + h) - f(9)}{h} = \frac{\frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}}{h} = 0, \ \forall h \neq 0[/tex]
    What does the above expression tell you?
  11. Jan 31, 2006 #10
    thanks, I got it now. Thanks a lot!
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