how do i get from this step of the limit: limit of [1/(3+x)-(1/3)]/x as x approaches 0 to this step limit of 3-(3+x)/(3+x)3(x) as x approaches 0 the solution to the problem is -1/9
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$\displaystyle \begin{gathered} \frac{1} {{3 + x}} - \frac{1} {3} = \frac{{ - x}} {{9 + 3x}} \hfill \\ \frac{{\frac{{ - x}} {{9 + 3x}}}} {x} = \frac{{ - 1}} {{9 + 3x}} \hfill \\ \end{gathered} $
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