Limits

First , transformed into an algebraic limit

Now using standard following theorems ,

$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$

We get ,

$\begin{aligned}L &= \lim_{x \to 0}\frac{(x + 4)^{3/2} + e^{x} - 9}{x}\\ &= \lim_{x \to 0}\frac{(x + 4)^{3/2} - 8}{x} + \frac{e^{x} - 1}{x}\\ &= \lim_{x \to 0}\frac{(x + 4)^{3/2} - 8}{(x + 4) - 4} + 1\\ &= \lim_{t \to 4}\frac{t^{3/2} - 4^{3/2}}{t - 4} + 1\text{ (by putting }t = x + 4)\\ &= \frac{3}{2}\cdot 4^{1/2} + 1 = 4\end{aligned}$

This approach is without L Hosp. rule. View @Svatejas Shivakumar 's solution for the bonus one.

$\begin{aligned} L & = \lim_{x \to 0} \frac{\color{#3D99F6}{(x+4)^{\frac{3}{2}}}+\color{#D61F06}{e^x}-9}{x} \\ & = \lim_{x \to 0} \frac{\color{#3D99F6}{8(1+\frac{x}{4})^{\frac{3}{2}}}+\color{#D61F06}{e^x}-9}{x} \quad \quad \small \color{#3D99F6}{\text{Using Maclaurin series}} \\ & = \lim_{x \to 0} \frac{\color{#3D99F6}{8\left(1+\frac{3x}{8}+\frac{3x^2}{128} - \frac{x^3}{1024}-... \right)}+\color{#D61F06}{\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+...\right)}-9}{x} \\ & = \lim_{x \to 0} \frac{\color{#3D99F6}{\left(8+3x+\frac{3x^2}{16} - \frac{x^3}{128}-... \right)}+\color{#D61F06}{\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+...\right)}-9}{x} \\ & = \lim_{x \to 0} \frac{4x + \frac{11}{16}x^2 + \frac{61}{384}x^3+...}{x} \\ & = \lim_{x \to 0}\space 4 + \frac{11}{16}x + \frac{61}{384}x^2+... \\ & = \boxed{4} \end{aligned}$

Substituting $x=0$ , $L$ takes the form of $\frac{0}{0}$ , the derivative of $L$ is $\frac{3}{2} \times (x+4)^{1/2}$ . By L'Hopital's Rule substituting the derivative of $L$ with $x=0$ we get the required limit as $\boxed{4}$