Limits practice question 15

Both $\sqrt{9x^{2} + 4x}$ and $3x$ grow without bound. Does that settle their difference, and what trick turns a root minus a term into a single fraction?

The form is $\infty - \infty$. Multiply and divide by the conjugate $\sqrt{9x^{2} + 4x} + 3x$.

After the difference of squares the numerator is $4x$. Divide the top and the bottom by $x$, using $\sqrt{9x^{2} + 4x} = x\sqrt{9 + \tfrac{4}{x}}$ for $x > 0$.