Question

How would I solve this limit problem? The answer isn’t zero though I put that and it’s wrong?

Answer 1

0 is correct, though...

`(\frac1x - \frac13)/(x + 3)`

is continuous at
`x=3`, so

`\displaystyle \lim_(x\to3) (\frac1x - \frac13)/(x + 3) = (\frac13 - \frac13)/(\frac13 + 3) = \frac0{\frac{10}3} = 0`

What was probably intended was the limit

`\displaystyle \lim_(x\to3) (\frac1x - \frac13)/(x - 3)`

which requires a little more finesse because the function is not continuous at
`x=3`.

Rewrite it as

`(\frac1x - \frac13)/(x - 3) = (3 - x)/(3x(x-3)) = -(x-3)/(3x(x-3))`

When
`x\\eq3`, we can cancel
`x-3` so that

`\displaystyle \lim_(x\to3) (\frac1x-\frac13)/(x-3) = \lim_(x\to3) -\frac1{3x} = -\frac1{3*3} = -\frac19`