**ELEMENTARY CALCULUS! Limits. Please help!?**

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*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!

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)

**Yahoo! Answers?**