Question

Evaluating Limits

Answer 1

\[\lim_{x \rightarrow 3}\frac{ x-3 }{ x ^{2}-9 }\]

Answer 2

hint factor the denominator

Answer 3

If I factor the denominator, I'm left with lim (as x approaches 3) of x+3 @freckles but what do I do with what's left?

Answer 4

right
recall x approaches 3

Answer 5

well actually it is 1/(x+3)

Answer 6

not x+3

Answer 7

@Sunshine447 have you learned l'hopital's rule yet?
@freckles refresh me, this seems to be in an indeterminate form. So if we factor out the denominator and plug in 3 for x, wouldn't that not accomplish anything?

Answer 8

1/(x+3) is continuous at x=3
therefore you can use direct sub @Jamierox4ev3r

Answer 9

I see. And that is a more elementary method. Nice

Answer 10

Yes that is the way usually thought before learning of l'hospital

Answer 11

but l'hospital is also nice

Answer 12

taught*

Answer 13

You can use difference of squares \[a^2-b^2 = (a+b)(a-b)\]

Answer 14

I personally like to use l'hopital as soon as I see infinity over infinity, or 0/0 lol

Answer 15

@Astrophysics I believe this was already done. By factoring the denominator

Answer 16

\[\lim_{x \rightarrow 3} \frac{ (x-3) }{ (x+3)(x-3) } = \lim_{x \rightarrow 3} \frac{ 1 }{ (x+3) }\]

Answer 17

Oh obviously I don't read, freckles already wrote that

Answer 18

@Sunshine447 do you have any questions?

Answer 19

if f(x) is continuous at x=a
then you can use direct sub
which I'm saying this:
\[\lim_{x \rightarrow a}f(x)=f(a) \text{ if } f(x) \text{ is continuous at } x=a\]