Expert: Alon Mandes - 9/1/2008

Question
QUESTION: Hello!

I am learning Calculus from the C.O.W on-line tutorial, just for fun and as a personal challenge , and although my mathematical background is not very strong, ( I did algebra in High School about 50 years ago..) I have brushed up my basics through a pre-calculus on-line tutorial and I am now, (quite unexpectedly !) able to grasp the conceptual side of Limit and to solve the problems of the tutorial.
I must still have, though, some loopholes in my algebra, because  I am now stuck with the following drill for an algebraic problem:

Find the limit of f(x)=x^3+3x^2 at x=near-3

By replacing x with (-2+h), I have to work with a cubic power expression, i.e. (-2+h)^3, and I don't know the trick  to expand it, as I would be able to ,if it were to the 2nd power!

Can you please provide a step-by-step solution of the problem, indicating how to expand a cubic power expression such as ( -2+h)^3 in order to arrive at the Limit ?

Thank you

Franco V.

ANSWER: Hi Franco, the math rule for the binom of third form is :
(a+b)^3=a^3+3a^2b+3ab^2+b^3.
I think you have been mistaken, it should be h-3 instead of h-2.
So, our form will become :
(h-3)^3=h^3-9h^2+27h+(-3)^3. Let's take a look at our limit :
Lim{x --> -3} of x^3+3x^2 = Lim{h --> 0} of (h-3)^3+3(h-3)^2
= Lim{h --> 0} of h^3-9h^2+27h-27+3(h^2-6h+9)
= Lim{h --> 0} of h^3-9h^2+27h-27+3h^2-18h+27
= Lim{h --> 0} of h^3-6h^2+9h
= 0

Hope This could help you. Any problems, I'm here.

Alon.

---------- FOLLOW-UP ----------

QUESTION: Thank you for your elucidation, Mr Alon.

Do I understand correctly that you get 0 as a limit because if you substitute x=-3 in  the expression f(x)=x^3+3x^2   you  get exactly  0 ? And/or is it also because if you plug in x=-3 in the last expression (h^3-6h^2+9h ) you also get  0  (27 -54+27)?

In its explanation of the Limit, the C.O.W. extracts the limit  after getting to  an expression  A + (h)times epsilon where A is a number (a constant) and h times epsilon is vanishingly small and goes to 0. In fact,  I remember another drill where the expression was in the form of , say, 17/32 + h^3-6h^2+9h  and the C.O.W. was considering the limit as 17/32. Am I getting the right rationale?
In the case I submitted,  there was no constant in the expression h^3-6h^2+9h , only h values, and the expression went to 0 by plugging in -3 for x. Is this why you arrive at a 0 limit ?

Thank you for your help

Franco

Answer
The expression A+(h)times epsilon in our case is :
0+h^3-6h^2+9h . Since h^3-6h^2+9h is vanishingly small and goes
to zero , we remain with 0. To chick our result, if it's true or
not, we plug -3 in the original function f(x) & we see that we
get the same result : 0.
We can't plug -3 in h^3-6h^2+9h, no meaning for this !!
The Issue of the limit consideration is to get to a form that is
compounded of a stable value + another diminishing form. This is
called : converge to the value.

Alon.