Question

Evaluate? `lim/(xrarr0)``((2+h)^3-8)/h`

Answer 1

`lim_(h->0)(((2+h)^3-8)/h)=12`

Explanation:

Hmm... this looks similar to the definition of a derivative:

`lim_(h->0)((f(x+h)-f(x))/h)`

In `((2+h)^3-8)/h`, we seem to be cubing a given value of `x`.
Therefore, we say that `f(x)=x^3`.

We can rewrite our problem:
`lim_(h->0)(((2+h)^3-8)/h)`

=>`lim_(h->0)(((2+h)^3-2^3)/h)`

Now, our problem seems to ask:
"What is the instantaneous rate of change of the function `f(x)=x^3` when `x=2`?"

Instead of trying out this limit, we can now calculate the derivative of the function `f(x)=x^3` when `x=2` using the power rule.

Power rule states that: If `f(x)=x^n` then `f'(x)=nx^(n-1)` where `n` is a constant.

We apply the rule to our situation:
Since `f(x)=x^3`,

`f'(x)=3x^2`

=>`f'(2)=3(2^2)`

=>`f'(2)=12`

That is our answer!