Hello.
I have to find the limit as x approaches 0 of the function
[ [(1/(3+x)] -(1/3) ] / x i tried direct substitution and to rationalize the denominator and numerator, but neither works...
what shoul i do?
$\displaystyle \frac{\frac{1}{3 + x} - \frac{1}{3}}{x} = \frac{\frac{3 - (3 + x)}{3(3 + x)}}{x}$
$\displaystyle = \frac{-\frac{x}{3(3 + x)}}{x}$
$\displaystyle = -\frac{1}{3(3 + x)}$.
So $\displaystyle \lim_{x \to 0}\left(\frac{\frac{1}{3 + x} - \frac{1}{3}}{x}\right) = \lim_{x \to 0}\left[-\frac{1}{3(3 + x)}\right]$
$\displaystyle = -\frac{1}{3(3)}$
$\displaystyle = -\frac{1}{9}$.