# Slope of tangent line

1. Feb 27, 2009

### louie3006

1. The problem statement, all variables and given/known data

find the slope of the tangent line whose g(x)=x^2-4 at point (1,-3)

2. Relevant equations

lim f(x+Δχ) -F(c)/ (Δχ)

3. The attempt at a solution
g(x)= x^2-4
G(1+Δχ)= (1+Δχ)^2-4 ==> Δχ^2+2Δχ-3

lim (Δχ^2+2Δχ-3) - (-3)/(Δχ) = 0
but in the book the answer is 2 so what could I've done wrong?

2. Feb 27, 2009

### Staff: Mentor

Try to keep your letters straight. You have g and G and f and F and x and X and c. There will come a time when this will get you in trouble.

Your last expression (which by the way isn't equal to 0), when simplified a bit, is
$[\Delta x ^2 + 2 \Delta x - 3 + 3]/\Delta x$

= $(\Delta x ^2 + 2 \Delta x)/\Delta x$

Factor $\Delta x$ from both terms in the numerator, and cancel with the one in the denominator, then take the limit as $\Delta x$ goes to zero.

3. Feb 27, 2009

### lanedance

the limt looks ok until you jump to 0, you still have a deltaX on the denominator, which would tend towrds infinty while the top will tend towards zero. so at teh moment you limit is undetermined until you clean it up a bit more...

so you need to cancel deltaX as much as possible before taking the limit

4. Feb 28, 2009

### HallsofIvy

Staff Emeritus
I assume you mean g(1+ Δx)

That limit is NOT 0. If you simply set Δx= 0 you get 0/0 so you have to be more careful.(Δχ^2+2Δχ-3) - (-3)= (Δx)^2+ 2Δx so [(Δχ^2+2Δχ-3) - (-3)]/(Δχ) = (Δx^2+ 2Δx)/Δx= Δx + 2. Take the limit, as Δx goes to 0 of Δx+ 2.