Worked example of a limit

1. Sep 16, 2008

naele

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

2. Relevant equations

$$m = \lim_{h\rightarrow 0}\frac{f(a+h) - f(a)}{h}$$

3. The attempt at a solution

His solution goes as such:

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

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

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

4) $$\lim_{h\rightarrow 0}-\frac{1}{3+h}=-\frac{1}{3}$$
I would have gotten 1/3 instead of -1/3 so I would have made a mistake between steps 2 and 3.

2. Sep 16, 2008

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.

Last edited: Sep 16, 2008
3. Sep 16, 2008

naele

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

4. Sep 16, 2008

HallsofIvy

Staff Emeritus
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!