Question

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

Answer 1

Add some fractions and do some algebra to find `lim_(x->0)(1/(3+x)-1/3)/x=-1/9`.

Explanation:

I'm assuming you mean `lim_(x->0)(1/(3+x)-1/3)/x`.

Direct substitution produces the indeterminate form `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->0)(1/(3+x)-1/3)/x`
`=(1/(3+x)-1/3((1+x/3)/(1+x/3)))/x`
`=(1/(3+x)-(1+x/3)/(3+x))/x`
`=((1-(1+x/3))/(3+x))/x`
`=((1-1-x/3)/(3+x))/x`
`=((-x/3)/(3+x))/x`
`=(-x)/(3(3+x)x)`
`=-1/(3(3+x)`

Phew! Now we can try direct substitution again:
`lim_(x->0)-1/(3(3+x)`
`=-1/(3(3+0))=-1/9`

And we're done.