L'Hospital's Rule works nicely in both cases. If you have an indeterminate form $\displaystyle \displaystyle \frac{0}{0}$ or $\displaystyle \displaystyle \frac{\infty}{\infty}$, then

$\displaystyle \displaystyle \lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}$

The first is already in $\displaystyle \displaystyle \frac{\infty}{\infty}$ form, and the second can be made into that form using this transformation...

$\displaystyle \displaystyle \begin{align*} \sqrt{9x^2 + x} - 3x &= \ln{\left(e^{\sqrt{9x^2 + x} - 3x}\right)} \\ &= \ln{\left(\frac{e^{\sqrt{9x^2 + x}}}{e^{3x}}\right)}\end{align*}$

Recall that because of the continuity of the logarithm, $\displaystyle \displaystyle \lim_{x \to c}\ln{[f(x)]} = \ln{\left[\lim_{x \to c}f(x)\right]}$, and the "stuff" inside the logarithm is now of the form $\displaystyle \displaystyle \frac{\infty}{\infty}$. So you can use L'Hospital's Rule there too