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

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: $\sqrt{x^2+9}+3$.

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 $$f'(0)=1/6$$

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.