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Worked example of a limit

  1. Sep 16, 2008 #1
    1. The problem statement, all variables and given/known data

    This is a worked example from Stewart's Early Transcendentals 6e section 2.7 p. 145 for anybody curious.

    Let [tex]f(x)= \frac{3}{x}[/tex]. Find an equation of the tangent line to the hyperbola at point (3,1).

    2. Relevant equations

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

    3. The attempt at a solution

    His solution goes as such:

    1) [tex]m = \lim_{h\rightarrow 0}\frac{f(3+h) - f(3)}{h}= \lim_{h\rightarrow 0}\frac{\frac{3}{3+h} -1}{h}[/tex]
    Plug in the point coordinates into the equation and evaluate.

    2) [tex]\lim_{h\rightarrow 0}\frac{\frac{3-(3+h)}{3+h}}{h}[/tex]
    Consider the 1 as 1/1, cross multiply and multiply through the denominator. The reverse of partial fraction decomposition (recomposition?)

    3) [tex]\lim_{h\rightarrow 0}\frac{-h}{h(3+h)}[/tex]
    This is where I become confused. Do you have to distribute the negative sign such that 3-(3+h) = 3-3-h = -h?

    4) [tex]\lim_{h\rightarrow 0}-\frac{1}{3+h}=-\frac{1}{3}[/tex]
    I would have gotten 1/3 instead of -1/3 so I would have made a mistake between steps 2 and 3.
     
  2. jcsd
  3. Sep 16, 2008 #2

    CompuChip

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    Would it help if I wrote it this way?
    3 - (3 + h) = 3 + (-1)(3 + h)

    Also, if you'd gotten 1/3 you knew you would be wrong, because the graph is clearly going down so the derivative should be negative.
     
    Last edited: Sep 16, 2008
  4. Sep 16, 2008 #3
    Yes! But then why did Stewart write it like that? It's so confusing.
     
  5. Sep 16, 2008 #4

    HallsofIvy

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    Perhaps Stewart didn't think he had to remind you that -(3-h)= -3+ h. You shouldn't have to ask if you use the distributive law!
     
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