How to solve a limit through rationalizing but the radicand is a linear relation and the denominator is a quadratic?

Question

The question is :

$$\lim_{x\to 0}{\frac{\sqrt {3x+4}-6}{x^2}}$$

My friend and I are absolutely stumped and can only attempt to solve through substitution. Any ideas?

Answer 1

For small x, -6 < $\sqrt {3x+4}$ - 6 < -1.
Since x$^2$ is positive, for small x
-6/x$^2$ < ($\sqrt {3x+4}$ - 6)/x$^2$ < -1/x$^2$.

As x approaches 0, the left hand and the right hand sides approach minus infinity, thus forcing the middle to approach minus infinity.

Answer 2

At the limit, the numerator is non-zero and negative and the denominator is zero.

What does that tell you?