Question

Limit Help. I am typing the equation

Answer 1

\[\lim_{x \rightarrow 3}\frac{ 6-4x ^{3}-96 }{ x-3 }\]

Answer 2

woops typed it wrong

Answer 3

perhaps we can factor out the x-3 ?

Answer 4

\to works for the rightarrow

Answer 5

\[\lim_{x \rightarrow 3}\frac{ 6-4x ^{3}-102 }{ x-3 }\]

Answer 6

Yea, I have tried to factor, but I cannot figure out how to get x-3 as a factor

Answer 7

does the top zero out when x=3?

Answer 8

not sure why you have 6-102 instead of just -96

Answer 9

the latex coding is not showing up on my computer, can you provide a screen shot?

Answer 10

well, its because i am tring to find the tangent line at x = 3 of \[f(x) = 6-4x ^{3}\]

Answer 11

the derivative of f(x) is just -12x^2, so the slope of the line at x=3 is just -108

Answer 12

therefore the tangent line is defined for some x=c as:

y = f'(c)(x-c)+f(c)

Answer 13

i spose youre having to do the derivative the long way?

Answer 14

Maybe, I need to reread the chapter. I assumed I could find the tangent at x=3 by finding the limit of the secant line. Hence the second equation i typed.

Answer 15

the derivative is defined as the limit as h approaches 0 of:

f(x+h) - f(x)
-----------
(x+h)-x


correct?

Answer 16

yep

Answer 17

then we can work it as:

6-4(x+h)^{3} - (6-4x^{3})
----------------------
h



(x+h)^{3} - x^{3}
-4 ------------------
h




x^3+3x^2h+3xh^2+h^3 - x^{3}
-4 -----------------------------
h



3x^2h+3xh^2+h^3
-4 ------------------
h



-4 (3x^2+3xh+h^2), when h=0 is just -4(3x^2), or simply -12x^2

Answer 18

as such, -12(3)^2 = -108

Answer 19

Ah, I guess I was just approaching the solution with the wrong approach. Thanks

Answer 20

youre welcome