How can the limit of a function be calculated using the standard formula?

[BayernBlues]

Homework Statement

http://img159.imageshack.us/img159/8161/64625334ex7.png

Homework Equations

The Attempt at a Solution

(f(a)+h)-f(a)
------------
h

f(a)=x^3 and f(a)=8
f(a)=2

I'm pretty sure that the answer of f(a)=2 is right but is there a better way to show it and I'm also unsure if I answered it fully.

 

[neutrino]

f(a) is not 2. The limit actually refers to the derivaive of the function at the point 2. Therefore, a=2.

 

[BayernBlues]

Okay; a=2

How do I show this though because if there's a test, the teacher isn't going to give me marks because I didn't show my work.

 

[theriel]

I would write:
f(a+h) = (2 + h)^3, hence a+h=2+h, hence a=2. Therefore f(x)=x^3. Assume it is correct and use the other part of the equation to confirm the assumption.

f(a)=8. We use previously found formula -> f(a)=a^3=8. We solve for a.
a=2 what holds according to the data provided.

 

[mjsd]

i guess it is a matter of comparing with the standard formula:
$$\lim_{h\rightarrow 0}\; \frac{f(x+h)-f(x)}{h}$$

so all I think you need to do is the following:
$$\lim_{h\rightarrow 0}\; \frac{(2+h)^3-8}{h} =<br /> \lim_{h\rightarrow 0}\; \frac{(2+h)^3-2^3}{h}<br /> = \left.\lim_{h\rightarrow 0}\; \frac{(x+h)^3-x^3}{h}\right|_{\text{at}\; x=2}$$
then you can identify what f(x) and x=a are.