Question

find limits if they exist f(x)={1+|3-x| if x is unequal to 3}
{2 if x=3}
a.) lim x----->3^+
b.)lim x----->3^-
c.)lim x ----> 3

Answer 1

@amriju

Answer 2

@uri
@John_ES

Answer 3

you sure you have the problem stated correctly?

Answer 4

Yes, I can reword it because it isn't stated the way it is in the problem

Answer 5

\[f(x)=\left\{ \frac{ 1+|3-x| if x \neq3 }{ 2 if x =3 } \right\}\]

Answer 6

DNE
3+ = 2
3 - = DNE
2 not equal to DNE, so the limit doesnot exists

Answer 7

For all 3 equations things then?

Answer 8

for X--->3= DNE
X can never be equal to 3 so 3^- DNE when X--->3

Answer 9

no...
limit as x ->3^+ = limit as x ->3^- = 2

Answer 10

Alright well how about this problem then

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

Answer 11

0

Answer 12

Yes its zero but how would I put the answer?

Answer 13

SO the complete equations for the first problem I asked would become DNE because of the 2 with x=3

Answer 14

yes

Answer 15

expand it for example sqrt (3-x)^2

Answer 16

no...

Answer 17

my computer is having issues. if your function is 1 + |3 - x| when x isn't 3 and 2 when x is 3 then the limit as x goes to 3 is 2.

Answer 18

(3-x)= 3-3 = 0

Answer 19

oops...
the limit of the first part is 1 and the second part is 2. it's not continuous but the limit does exist and the limit is 1.

Answer 20

the first part doesnt exist since X can not be 3

Answer 21

f(3) = 2 but limit f(x) as x -> 3 = 1.

Answer 22

You guys are confusing me.

Answer 23

they are not equal to eachothe therefore, X-> 3= DNE

Answer 24

Thank you! And then for the second equation we wouldn't put the lim is 0. Wouldn't that be the indeterminate

Answer 25

is this your function?\[f(x) =\left[\begin{matrix}1+|3-x| & x\neq 3 \\ 2& x=3\end{matrix}\right] \]

Answer 26

Yes it is

Answer 27

and you want this?\[\lim_{x \rightarrow 3} f(x)\]

Answer 28

No I want.