Finding the Tangent Line to a Hyperbola: A Worked Example Using Limits

[naele]

Homework Statement

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).

Homework Equations

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

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.

 

[CompuChip]

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.

 

[naele]

Yes! But then why did Stewart write it like that? It's so confusing.

 

[HallsofIvy]

Yes! But then why did Stewart write it like that? It's so confusing.

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!