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

Question

asked Jan 3, 2023 7.8k views

1 Answer

Sschmeck · Answer 1 · 2023-01-07T22:27:46+0000

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

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