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Another equation of a tangent

  • Thread starter teneleven
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  • #1
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Homework Statement



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

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


Homework Equations



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

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


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:

Answers and Replies

  • #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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