How to find the limit of $\frac{\sqrt{x^2+9}-3}{x^2}$? as x approaches 0 
How to find the limit of $\frac{\sqrt{x^2+9}-3}{x^2}$? as x approaches 0

I thought that this was going to be a simple problem but then it got more complicated than I have expected
So I first multiplied the expression with its conjugate on both the denominator and the numerator
the conjugate was the following:
$$\sqrt{x^2+9}+3$$
but when I multiplied by the conjugate I actually didn't even get close to the answer because the answer was supposed to be $\frac{1}{6}$ 
instead I have gotten the following by multiplying by its conjugate
$\frac{x^2+9-9}{x^2\sqrt{x^2+9}-3x^2}$ which obviously doesn't return 1/6
I was wondering if I was just simply making a mistake or if I have to choose a different option in solving this?

Note:
\begin{align*}
\frac{\sqrt{x^2+9}-3}{x^2}
= & \; \frac{\big(\sqrt{x^2+9}-3 \big) \big(\sqrt{x^2+9}+3 \big)}{x^2 \big( \sqrt{x^2+9}+3 \big)} \\
= & \; \frac{x^2+9-3^2}{x^2\big(\sqrt{x^2+9}+3 \big)} \\
= & \; \frac{1}{\sqrt{x^2+9} + 3}.
\end{align*}
So we have
\begin{align*}
\lim_{x \to 0} \frac{\sqrt{x^2+9}-3}{x^2} = \lim_{x \to 0} \frac{1}{\sqrt{x^2+9} + 3} = \frac{1}{\sqrt{9}+3}=\frac{1}{6}.
\end{align*}
I guess you made some mistake when you multiplied the conjugate.
