<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] 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 $x=0$.
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