Question

limits...

Answer 1

Tried anything yet?

Answer 2

Find the limit when x approaches 9^- for x+9, and find the limit x approaches 9^+ and see if it exists or not.

Answer 3

wait wouldnt i plug in 9?

Answer 4

it would be 18 but since x<9 then wouldnt it not exist?

Answer 5

Do left and right side to see if it exists

Answer 6

well 27-9=18 also

Answer 7

\[\huge \lim_{x \rightarrow 9^{-}} x+9\]


\[\huge \lim_{x \rightarrow 9^{+}} 27-x\]

Answer 8

So does the limit x-> 9 exist? :P

Answer 9

If they both = 18, yes the limit is continuous and it exists.

Answer 10

wait @iambatman so for this one
the limit wouldnt exist since one of them =0?

Answer 11

i only need some one to check if its right .-.

Answer 12

Same process, what did you get?

Answer 13

i already sadi, i believe that the limit does not exist

Answer 14

I believe you're right

Answer 15

Go my batman, go ;)

Answer 16

yay c: i have like 3 more c: if you will please....

Answer 17

so itd be 1/9-9 which is 1/0 so its dosent exist c:

Answer 18

UNDEFINED, yup.

Answer 19

oh just kidding i have to find vertical asymptotes .-.

Answer 20

NOOOOOOOOOOOOO

Answer 21

and identify the limit... so the limits -infinity and vertical.... i have no idea .-.

Answer 22

You have to look at the denominator of w e you just showed me... and what makes it = 0 is a candidate for vertical asymptote, so in your case 9. Then you have to find the limit and see if it = infinity or not, if it does, than it will have a vertical asymptote.

Answer 23

wait so the vertical asymptote in 9 and the limits -infinity right?

Answer 24

@Luigi0210

Answer 25

Well here's what your graph looks like:

Answer 26

\(\Large \lim_{x \rightarrow 9^+} f(x)=\infty\) and \(\Large \lim_{x \rightarrow 9^-} f(x)=-\infty\) so in conclusion \(\Large \lim_{x \rightarrow 9} f(x)=DNE\) because \(\Large \lim_{x \rightarrow 9^+} f(x) \neq \lim_{x \rightarrow 9^-} f(x) \)