Compute $\displaystyle \lim_{x \rightarrow -3} \frac{\frac{1}{3} + \frac{1}{x}}{3+x}$ $$\lim_{x \rightarrow -3} \frac{\frac{1}{3} + \frac{1}{x}}{3+x}$$
Can I multiply the top and bottom by $3x$? If not, why not. I am having a hard time understanding when it is appropriate to multiply the numerator and denominator by the LCD. 

\begin{align*}\lim_{x\rightarrow -3}\ \frac{\frac{1}{3} + \frac{1}{x}}{3+x} & = \lim_{x\rightarrow -3}\ \frac{x+3}{3x(3+x)} \\&= \lim_{x\rightarrow -3}\ \frac{1}{3x} = \frac{-1}{9}\end{align*}
