<s>[INST] 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! 
 [/INST] 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 $\;(3-\sqrt{x+9})(3+\sqrt{x+9})=9-(x+9)=\color{red}{-x}\;$  **
$$=(-1)\frac{3+\sqrt9}{\sqrt{49}+7}=-\frac6{14}=-\frac37$$
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