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Homework Help: Relating limits to derivatives, as x approaches non zero number?

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose that f' (2) = 3. Find the limit as x approaches 2 of [f(x)−f(2)]/[sqrt(x) - sqrt(2)]

    Answer: 6*sqrt 2

    2. Relevant equations

    3. The attempt at a solution

    f'(x) = lim h->0 = [f(a +h) - f(a)]/h = slope [f(x)-f(2)]/ x-2

    a = 2

    i would think that the limit = 3? wrong answer

    tried graphing it, but no original function is given
    sqrtx and sqrt2 confuses me and not sure how to do it.

    thank you
  2. jcsd
  3. May 22, 2013 #2


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    Rationalize the denominator of ##\displaystyle \ \frac{f(x)−f(2)}{\sqrt{x\,}-\sqrt{2\,}}\ .##
  4. May 22, 2013 #3


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    Equivalently, write
    [tex]\frac{f(x)- f(2)}{\sqrt{x}- \sqrt{2}}= \frac{f(x)- f(2)}{x- 2}\frac{x- 2}{\sqrt{x}- \sqrt{2}}[/tex]
    [tex]= \frac{f(x)- f(2)}{x- 2}\left(\frac{1}{\frac{\sqrt{x}- \sqrt{2}}{x- 2}}\right)

    \text{ and think about the derivatives of f(2) and }\sqrt{x}[/tex] at x= 2.
    Last edited by a moderator: May 22, 2013
  5. May 22, 2013 #4
    I think i've got it :) how did you create those math symbols and format it that way? was it through the physicsforums website coding or by a program?

    f'(2) = 3
    x = 2

    lim x->2 [f(x) - f(2)]/(x - 2) * lim x->2 1/ [sqrtx - sqrt2]/(x-2)

    = (3) * 1/f'(sqrtx)
    = 3 * 1/(1/2sqrtx)
    = 3 * 2sqrtx
    = 3 * 2 sqrt2
    = 6 sqrt2
  6. May 22, 2013 #5

    Simon Bridge

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    It's called ##\LaTeX## ;)

    Underneath each post you will see a button marked "quote" - if you click on it, you will be taken to an advanced editor screen where everything in the post is included in "quote" tags. Read what is in the quote to see how all the symbols came out nicely formatted ;)

    Use double-hash or double-dollar signs around the block of math you want formatted - and use a backslash "\" in front of the bits that need special formatting - so \sqrt{ \frac{\alpha}{\beta} } will come out $$\sqrt{\frac{\alpha}{\beta}}$$
  7. May 22, 2013 #6


    Staff: Mentor

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