How do you find the limit of (1/(3+x)-(1/3))/(x) as x->0?

Nov 25, 2016

${\lim}_{x \rightarrow 0} \frac{\frac{1}{3 + x} - \frac{1}{3}}{x} = - \frac{1}{9}$

Explanation:

${\lim}_{x \rightarrow 0} \frac{\frac{1}{3 + x} - \frac{1}{3}}{x}$

We can clear this limit of fractions by multiplying as follows:

$= {\lim}_{x \rightarrow 0} \frac{\left(\frac{1}{3 + x} - \frac{1}{3}\right) \left(3\right) \left(3 + x\right)}{x \left(3\right) \left(3 + x\right)}$

Multiplying through in the numerator gives:

$= {\lim}_{x \rightarrow 0} \frac{\frac{3 \left(3 + x\right)}{3 + x} - \frac{3 \left(3 + x\right)}{3}}{3 x \left(x + 3\right)}$

Canceling:

$= {\lim}_{x \rightarrow 0} \frac{3 - \left(3 + x\right)}{3 x \left(x + 3\right)}$

$= {\lim}_{x \rightarrow 0} \frac{3 - 3 - x}{3 x \left(x + 3\right)}$

$= {\lim}_{x \rightarrow 0} \frac{- x}{3 x \left(x + 3\right)}$

$= {\lim}_{x \rightarrow 0} \frac{- 1}{3 \left(x + 3\right)}$

Now we can evaluate the limit, since there is no longer an issue for $x \rightarrow 0$.

$= \frac{- 1}{3 \left(0 + 3\right)}$

$= - \frac{1}{9}$