Question

[(1/(3+x))-(1/3)] / x Limit 0

Answer 1

First find a common denominator and combine the fractions in the numerator:

$\displaystyle\lim_{x\to0}\frac{\dfrac1{3+x}-\dfrac13}x=\lim_{x\to0}\frac{\dfrac3{3(3+x)}-\dfrac{3+x}{3(3+x)}}x=\lim_{x\to0}\frac{3-(3+x)}{3x(3+x)}$

Now simplify and cancel out all the terms that you can:

$\displaystyle\lim_{x\to0}\frac{3-3-x}{3x(3+x)}=-\frac13\lim_{x\to0}\frac1{3+x}$

Since the remaining expression is continuous as a function of $x$, you can directly substitute to end up with

$\displaystyle-\frac13\lim_{x\to0}\frac1{3+x}=-\frac13\times\frac1{3+0}=-\frac19$