Be careful! Evaluating this limit as $n → 0$ leads to trouble:

$\lim \limits_{n\rightarrow 0}\frac{\int_{2}^{2+0} \left(\sqrt{x^{3}+1}\right)dx}{n}=\frac{0}{0}\text{ Recall that }\frac{0}{0}\text{ is an indeterminate form.}$

That means we do not know if the limit is finite (because the $0$s 'cancel out') or infinite (because there is a $0$ in the denominator).

We can evaluate an indeterminate limit if that limit happens to be a definition of the derivative.
Sure enough, part (a) is Ana's Definition of the Derivative:

$f'(a)=\lim \limits_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$

Since $n → 0$, notice that $a = 0$. So we can rewrite the limit as:

$f'(0)=\lim \limits_{n\rightarrow 0}\frac{\int_{2}^{2+0}\left(\sqrt{x^{3}+1}\right)dx-0}{n-0}$

$\text{This means that }f(x)=\int_{2}^{2+n}\left(\sqrt{x^{3}+1}\right)dx.$

So, we just need to find $f '\left(x\right)$ at $x = 0$ and we have evaluated the limit.

$\text{By the Fundamental Theorem of Calculus: }\frac{d}{dx}\int_{2}^{2+n}\left (\sqrt{x^{3}+1} \right )dx=\sqrt{x^{3}+1}$

Now evaluate that derivative at $x = 0$:

$f'(0)=\sqrt{(0)^{3}+1}=1$

Refer to part (a).

$\lim \limits_{n\rightarrow -\infty}\frac{x^2}{e^{-x}}=\frac{\infty}{\infty}$

l’Hôpital’s Rule can be used.

See part (c).
Be sure to reevaluate the limit after each use of l’Hôpital’s Rule.