Limit with fraction in numerator

$\begingroup$

I need to find the limit for $$ \lim_{x\,\rightarrow\,0} \frac {\frac{1}{x+3} - \frac{1}{3}}{x} $$

It's supposed to be $-1/9$. I've tried changing it around multiple ways, and get the $9$ but never the $1$ over neg $9$.

$\endgroup$ 5

2 Answers

$\begingroup$

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

And you can continue from there.

$\endgroup$ $\begingroup$

The solution presented by @wltrup is solid and efficient. I thought that it would be instructive to present another way forward.

Here, straightforward application of L'Hospital's Rule reveals

$$\begin{align} \lim_{x\to 0}\frac{\frac{1}{x+3}-\frac13}{x}&=\lim_{x\to 0}\left(-\frac{1}{(x+3)^2}\right)\\\\ &=-\frac19 \end{align}$$

as expected!

$\endgroup$