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?
 A: Perhaps you made a slight mistake when jumping from one step to the next. I have hidden the work, and tried my best to hide the solution. For some reason, you can't hide a block of text and the indented $\LaTeX$. If you're stuck, just place your mouse over the yellow tabs and the work should show up.

Multiplying by the conjugate is the right thing to do! Let's multiply the fraction by $\sqrt{x^2+9}+3$. Therefore

$$\lim\limits_{x\to0}\frac {\sqrt{x^2+9}-3}{x^2}=\lim\limits_{x\to0}\frac {(\sqrt{x^2+9}-3)(\sqrt{x^2+9}+3)}{x^2(\sqrt{x^2+9}+3)}=\lim\limits_{x\to0}\frac {x^2+9-9}{x^2(\sqrt{x^2+9}+3)}$$

The numerator reduces down to $x^2$, and we see that both $x^2$ terms cancel each other out. Can you finish the rest?

$$\lim\limits_{x\to0}\frac {1}{\sqrt{x^2+9}+3}=\frac 1{\sqrt9+3}=\frac 16$$


Edit: The OP has provided his work on the problem. The denominator is actually supposed to be $3x^2$ and not $-3x^2$ because we're multiplying by the conjugate: $\sqrt{x^2+9}+3$.
A: With $f (x)=\sqrt {x+9} $, the limit is
$$\lim_0\frac {f (x)-f (0)}{x}=f'(0) $$
and $$f'(x)=\frac {1}{2f (x)} $$
thus $$f'(0)=1/6$$
A: 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.
A: Let $f(t) = \sqrt{t}$.
\begin{align*}
\text{Then}\;\;f'(9)&=\lim_{h \to 0}\frac{\sqrt{9+h}-\sqrt{9}}{h}\\[4pt]
&=\lim_{h \to 0^{+}}\frac{\sqrt{9+h}-\sqrt{9}}{h}\\[4pt]
&=\lim_{x \to 0}\frac{\sqrt{9+x^2}-3}{x^2}\\[4pt]
\end{align*}
A: Since both the numerator and denominator go to 0, it is undefined. I would suggest using L'hospitals rule, which states that if there is an indeterminate form $\frac{0}{0}$ you can differentiate the numerator and the denominator and take the limit of the result. Here we have $\frac{\sqrt{x^2+9}-3}{x^2}$. Once we differentiate the numerator and denominator and simplify we obtain $\frac{1}{2\sqrt{x^2+9}}$. If we take the limit as $x$ goes to $0$, we get that the limit is equal to $\frac{1}{6}$.
A: $$
\lim_{x\to0}\frac{\sqrt{x^2+9}-3}{x^2} = \lim_{x\to0}\frac{\frac{2x}{2\sqrt{x^2+9}}}{2x} = \frac{1}{2\cdot 3} = \frac{1}{6}
$$
