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{\frac1{3+x}-\frac13}x=\lim_(x\to0)\frac{\frac3{3(3+x)}-(3+x)/(3(3+x))}x=\lim_(x\to0)(3-(3+x))/(3x(3+x))`

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


`\displaystyle\lim_(x\to0)(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*\frac1{3+0}=-\frac19`