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! 
 A: 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.
A: You did the double multiplication together. That invites mistakes:
$$\lim_{x\to 0}  \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}=\lim_{x\to 0}  \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}\frac{\sqrt{x+49}+7}{\sqrt{x+49}+7}=$$
$$=\lim_{x\to 0}\frac x{\left(\sqrt{x+49}+7\right)\left(3-\sqrt{x+9}\right)}\frac{3+\sqrt{x+9}}{3+\sqrt{x+9}}=\lim_{x\to 0}\frac x{-x}\cdot\frac{3+\sqrt{x+9}}{\sqrt{x+49}+7}=$$
** because $\;(3-\sqrt{x+9})(3+\sqrt{x+9})=9-(x+9)=\color{red}{-x}\;$  **
$$=(-1)\frac{3+\sqrt9}{\sqrt{49}+7}=-\frac6{14}=-\frac37$$
A: At $x=0$, both numerator and denominator are $0$, so one way to do this is to apply L'Hospital's rule :
$$
\lim_{x\to 0} \frac{\sqrt{x+49}-7}{3-\sqrt{x+9}} = \lim_{x\to 0} \frac{1/2}{-1/2}\frac{(x+49)^{-1/2}}{(x+9)^{-1/2}}
$$
Now just plug in $x=0$.
A: You can also write your expression as
$$\frac{7}{3}\frac{\sqrt{1+x/49}-1}{1-\sqrt{1+x/9}}$$
If $x$ is small, then $\sqrt{1+x/49}=1+x/98, \sqrt{1+x/9}=1+x/18$
and finally we have $-3/7.$
