Question

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

Answer 1

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

Explanation:

`(1/(x+3)-(1/3))/x`

What do I not like about this?
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.)

`(1/(x+3)-(1/3)) = 3/(3(x+3))-(x+3)/(3(x+3))` (did that help? -- we're not done!)

`= (3-(x+3))/(3(x+3))` `" "` (still not done)

`= (-x)/(3(x+3))`

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

`(1/(x+3)-(1/3))/x = ((-x)/(3(x+3)))/x = ((-x)/(3(x+3)))/(x/1)`

`= ((-x)/(3(x+3)))* 1/x`

`= (-1)/(3(x+3))`

And now we can find the limit as `xrarr0`

`lim_(xrarr0) (1/(x+3)-(1/3))/x = lim_(xrarr0) (-1)/(3(x+3))`

`= (-1)/(3(0+3)) = (-1)/9`