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

Feb 19, 2017

Don't think too much about what you should do. Think a lot about what you could do.

#### Explanation:

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

One thing is the fraction in the numerator of the fraction.

What could I do about it?
Write the top as a single fraction. (Then we'll have a fraction over $x$.)

Will that help?

Well, we have to try something, so let's do it and if it doesn't help, we'll start over. (It will help.)

$\left(\frac{1}{x + 3} - \left(\frac{1}{3}\right)\right) = \frac{3}{3 \left(x + 3\right)} - \frac{x + 3}{3 \left(x + 3\right)}$ (did that help? -- we're not done!)

$= \frac{3 - \left(x + 3\right)}{3 \left(x + 3\right)}$ $\text{ }$ (still not done)

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

Let's see what the big fraction looks like now.

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

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

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

And now we can find the limit as $x \rightarrow 0$

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

$= \frac{- 1}{3 \left(0 + 3\right)} = \frac{- 1}{9}$