# How do you find the limit of [1/(3+x)]- (1/3) ÷ x as x approaches 0?

May 25, 2016

Add some fractions and do some algebra to find ${\lim}_{x \to 0} \frac{\frac{1}{3 + x} - \frac{1}{3}}{x} = - \frac{1}{9}$.

#### Explanation:

I'm assuming you mean ${\lim}_{x \to 0} \frac{\frac{1}{3 + x} - \frac{1}{3}}{x}$.

Direct substitution produces the indeterminate form $\frac{0}{0}$, and that doesn't help us very much.

Let's see what happens when we add the fractions in the numerator:
${\lim}_{x \to 0} \frac{\frac{1}{3 + x} - \frac{1}{3}}{x}$
$= \frac{\frac{1}{3 + x} - \frac{1}{3} \left(\frac{1 + \frac{x}{3}}{1 + \frac{x}{3}}\right)}{x}$
$= \frac{\frac{1}{3 + x} - \frac{1 + \frac{x}{3}}{3 + x}}{x}$
$= \frac{\frac{1 - \left(1 + \frac{x}{3}\right)}{3 + x}}{x}$
$= \frac{\frac{1 - 1 - \frac{x}{3}}{3 + x}}{x}$
$= \frac{\frac{- \frac{x}{3}}{3 + x}}{x}$
$= \frac{- x}{3 \left(3 + x\right) x}$
=-1/(3(3+x)

Phew! Now we can try direct substitution again:
lim_(x->0)-1/(3(3+x)
$= - \frac{1}{3 \left(3 + 0\right)} = - \frac{1}{9}$

And we're done.