ELEMENTARY CALCULUS! Limits. Please help!?

Question

ax^3-cx+7 = equation

a=7 c=21

Find the following limit.
lim h -> 0

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

1. First find the Simplified Difference Quotient (SDQ):

2. Now find the following limit:
lim h->0 SDQ=


This is my first week in Calculus. I have worked forever on these problems and cannot figure it out. I plan to get a math tutor ASAP, but until then.. I have hope with yahoo answers. Thanks guys!

Answer 1

Don't panic.

For f(x+h) simply replace x with x+h in the equation f(x) = ax^3 - cx +7.

This gives you:

f(x+h) = a*(x+h)^3 -c*(x+h)+7

f(x+h) = a*(x^3 +3x^2*h+3x*h^2+h^3)-cx -ch+7

now f(x+h) - f(x) = a*(x^3 +3x^2*h+3x*h^2+h^3)-cx -ch+7 - ax^3 + cx -7

or

f(x+h)-f(x) = 3ax^2*h+3ax*h^2+a*h^3-ch

Now divide by h

[f(x+h)-f(x)]/h = 3ax^2+3ax*h +ah^2 - c

Now take lim as h->0 = 3ax^2 - c = 21*(x^2-1)