Looking at the question, if the value of h = 0 is substituted directly into the question, we will obtain an indeterminate form
Method 1
L'Hôpital's Rule can help us calculate a limit that may otherwise be hard or impossible.
So, L’Hospital’s Rule tells us that if we have an indeterminate form 0/0 or
all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
[tex]\begin{gathered} \frac{\frac{d}{dx}((2+h)^3-8)}{\frac{d}{dx}(h)} \\ \\ \Rightarrow\frac{3(2+h)^2}{1} \end{gathered}[/tex]Then we can now put h = 0
[tex]3(2+0)^2[/tex]=> 3 x 4
=> 12
The answer = 12
Method 2
We can expand the numerator and then divide it by the denominator
[tex]\begin{gathered} \frac{(2+h)^3-8}{h} \\ \\ \frac{8+12h+6h^2+h^3\text{ - 8}}{h} \end{gathered}[/tex][tex]\frac{12h+6h^2+h^3}{h}[/tex][tex]12+6h+h^2[/tex]Substituting the value of h = 0
gives 12
