Will somebody look over my real-analysis/calculus solution of $\lim\limits_{x\to 0}\frac{1}{x} \int_0^{x} \sqrt{9+t^2}\mathrm{d}t$ Evaluate $\lim_{x\to 0}\frac{1}{x} \int_0^{x} \sqrt{9+t^2}\mathrm{d}t$.
Immediately, taking,
$$\begin{align}
&\lim_{x\to 0}\frac{1}{x} \int_0^{x} \sqrt{9+t^2}\mathrm{d}t \\
=&\frac{\lim_{x\to 0} \int_0^{x} \sqrt{9+t^2}\mathrm{d}t}{ \lim_{x\to 0}\frac{1}{x}}
\end{align}$$
which yields the indeterminate form $\frac{0}{0}$. 
Recall the

Fundamental Theorem of Calculus (1):   If $f$ is Riemann integrable on $[a,b]$ for $a\leq x\leq b$, put $F(x)=\int_0^{x}f(t)\mathrm{d}t$. Then $F$ is continuous on $[a,b]$: furthermore, if $f$ is continuous at a point $x_0$ of $[a,b]$, then $F$ is differentiable at $x_0$, and $F'(x)=f(x)$. 

From this point, using L'Hospital's Rule yields
$$\begin{align} 
&\lim_{x\to 0} \frac{1}{\frac{d}{dx}x} \cdot \lim_{x\to 0}\frac{d}{dx}\int_0^{x}\sqrt{9+x^2}\mathrm{d}t\\
\rightarrow& 1 \cdot \lim_{x\to 0}F'(x)\\
\rightarrow& \lim_{x\to 0} \sqrt{9+x^2}\\
=& 3.
\end{align}$$\
I feel like I have assumed something that I shouldn't have. Could somebody look this over for me?

Your answer is correct but you do not need to use L'Hospital's rule.
Let $F(x)=\int_{0}^x \sqrt{t^2+9}dt$.
 Then you have:
$$\lim_{x\rightarrow 0} \dfrac{F(x)-F(0)}{x-0}=F^\prime(0)$$
by definition of the derivative and the FTC gives you that $F^\prime(t)=\sqrt{t^2+9}$.
