can someone explain this limit i have,
$$\lim_{x\to 0}  \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$$ the correct answer is $-\frac{3}{7}$ and in my case the result is $\frac{7}{3}$ i don't understand. 
i tried this
$\lim_{x\to 0}  \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$=$\lim_{x\to 0}  \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {\sqrt{x+49}+7}{\sqrt{x+49}+7}.\frac {3+\sqrt{x+9}}{3+\sqrt{x+9}}$=
$\lim_{x\to 0}  \frac {x+49-49}{9-x+9}.\frac {\sqrt{x+49}+7}{3+\sqrt{x+9}}= \frac {x(\sqrt{x+49}-7)}{x(3-\sqrt{x+9})}=\frac{14}{3}=\frac{7}{3}$ i don't know what is the error! 

You made a couple mistakes. Observe that:
$$ \begin{align*}
\lim_{x\to 0}  \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {\sqrt{x+49}+7}{\sqrt{x+49}+7}.\frac {3+\sqrt{x+9}}{3+\sqrt{x+9}} 
&= \lim_{x\to 0}  \frac {(x+49)-49}{9-(x+9)}.\frac {3+\sqrt{x+9}}{\sqrt{x+49}+7} \\
&= \lim_{x\to 0}  \frac {x}{-x}.\frac {3+\sqrt{x+9}}{\sqrt{x+49}+7} \\
&= \lim_{x\to 0}  -\frac {3+\sqrt{x+9}}{\sqrt{x+49}+7} \\
&= -\frac {3+\sqrt{0+9}}{\sqrt{0+49}+7} \\
&= -\frac {6}{14} \\
&= \frac{-3}{7}
\end{align*}$$
as desired.
