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?

There is nothing wrong with the solution per se, but there are some errors in the presentation. When you invoke L'Hospital's rule, you write $\frac{d}{dx}\frac{1}{x}$ when you mean $\frac{1}{\frac{d}{dx} x},$ and also note that $F'(x) = \sqrt{9 + x^2}$, not $\sqrt{9+t^2}$, but other than that it's just fine!
