1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Another equation of a tangent
  1. Tangent's equation (Replies: 6)

Loading...