First principles-do both equations always work?

[idlackage]

First principles--do both equations always work?

Homework Statement

Find the slope of the tangent line to f(x) = 4 / (3-x) at the point x = 5.

Homework Equations

m = lim(x->a) f(x) - f(a) / (x-a)
m = lim(h->0) f(a+h) - f(a) / h

The Attempt at a Solution

With the second equation:

As the picture says, 1 is the correct answer for this question, but getting to it with the second equation brings me to the third last step which I'm not completely sure how to simplify. And by my guess, I'm doing it wrong. I can work it out fine with the first equation.

Semi relevant question--are both equations supposed to always work?

 

[tiny-tim]

hi idlackage!

your 4 - 2 - 2h on line 5 should be 4 - 4 - 2h

 

[eumyang]

I see a mistake between lines 5 and 6 (counting the original problem as line 1).

EDIT: Beaten to it.

 

[tiny-tim]

EDIT: Beaten to it.

only by h

 

[idlackage]

hi idlackage!

your 4 - 2 - 2h on line 5 should be 4 - 4 - 2h

Thank you! So now I can work it out as

* If I don't multiply and just cross out the h's, the denominator's -h canceled out by the numerator's +h--I can just ignore the signs? I always thought the top's leftover -2 would become positive or something, oopes. Do I just always cross out the numerator h with the 1/h's h, or do I assess the problem before deciding which to get rid of?

 

[Mark44]

"Crossing" out the h's is a very sloppy way to describe what you're doing, which is actually pulling out a factor of 1.

The underlying idea is that lim f(h)g(h) = lim f(h) * lim g(h), provided that the two separate limits both exist. In your problem you have in the penultimate step
lim (h/h)(-2/(-2 - h)) = lim (h/h)(2/(2 + h), and this can be written as the product of two limits lim (h/h) * lim 2/(2 + h). (The limits here should be understood as h --> 0.)

For any value of h other than zero, h/h = 1, so as h --> 0, h/h --> 1. The value of the other limit is also 1, so the value of the original limit expression is 1 as well.