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Homework Help: Another equation of a tangent

  1. Apr 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Find an equation of the tangent line to the curve at the given point.

    [tex]y = \frac{x - 1}{x - 2}[/tex] , (3, 2)


    2. Relevant equations

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

    [tex]y = mx + b[/tex]


    3. The attempt at a solution

    [tex]\lim_{h\to 0} \frac{\frac{(x + h) - 1}{(x + h) - 2} - \frac{x - 1}{x - 2}}{h}[/tex]

    [tex]\lim_{h\to 0} \frac{\frac{2 + h}{1 + h} - \frac{2}{1}}{h}[/tex]

    [tex]\lim_{h\to 0} \frac{2 + h}{h + h^2} - \frac{2}{h}[/tex]

    [tex]\lim_{h\to 0} \frac{2 + h}{h + h^2} - \frac{2 + 2h}{h + h^2}[/tex]

    [tex]\lim_{h\to 0} \frac{h + 2h}{h + h^2}[/tex]


    I'm not sure what to do after this step. The final answer is [tex]y = -x + 5[/tex]
    Thanks.
     
    Last edited: Apr 7, 2007
  2. jcsd
  3. Apr 7, 2007 #2

    arildno

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    Your last line is wrong (remember to change + to -!!).
    We have:
    [tex]\lim_{h\to{0}}(\frac{2+h}{h(1+h)}-\frac{2}{h})=\lim_{h\to{0}}\frac{2+h-2(1+h)}{h(1+h)}=\lim_{h\to{0}}\frac{-h}{h(1+h)}=\lim_{h\to{0}}\frac{-1}{(1+h)}=-1[/tex]
     
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