Question

Use the graph below to list the x value(s) where the limits as x approaches from the left and right of those integer values(s) are not equal.

Answer 1

@CausticSyndicalist @kohai @SolomonZelman @surjithayer @TheSmartOne

Answer 2

@SolomonZelman @surjithayer @NoelGreco @KMcc @kohai @Ridoru

Answer 3

Please help!

Answer 4

all graph is one function, correct?

Answer 5

by the limit being not equal, I will interpret as limit doesn't exist.

Answer 6

So for any x=a.
For the (2 sided) limit to exist

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~a^-}f(x)=\lim_{x \rightarrow ~a^+}f(x)}\) (one of the conditions of continuity)
~~~~~~~~~~~~~~~~~~~~~~~~~

IF, \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~a^-}f(x)~\ne~\lim_{x \rightarrow ~a^+}f(x)}\)

THEN, \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~a}f(x)}\) Does Not Exist.

Answer 7

there is one integer value, (looking at the graph) at which \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~a}f(x)}\) doesn't exist.

Answer 8

when x=-1?

Answer 9

at x=-1, all you have is a jump discontinuity. no.

Answer 10

shift discontinuity, whatever you want to refer to it as.

Answer 11

but there, \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-1^{+}}f(x)=\lim_{x \rightarrow ~-1^{+}}f(x)~~\Rightarrow~~~~y=-1}\)

Answer 12

at x=3 from both the left and the right they are both approaching -infinity

Answer 13

I meant
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~-1^{-}}f(x)=\lim_{x \rightarrow ~-1^{+}}f(x)~~\Rightarrow~~~~y=-1}\)

Answer 14

yes, so that means that \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~3^{-}}f(x)=\lim_{x \rightarrow ~3^{+}}f(x)}\) right?

Answer 15

and that means the limit \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~3}f(x)}\) does exist or does not ?

Answer 16

it exists because both sides equal one another

Answer 17

Yes.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
maybe there is some confusion about what I said. Like you meant one thing and said the other. But I replied only to what you said but not meant.

you asked me that is it at x=-1 that the limit doesn't exist. (right ?)
but I think you can see that the limit does exist at x=-1, despite the jump discontinuity.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There is that one value where the limits equal different values.
You can simply see this from the graph....

Answer 18

hint: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~1^{+}}f(x)=?}\)

Answer 19

1?

Answer 20

@SolomonZelman

Answer 21

1 from the right is 2 & 1 from the left is -1

Answer 22

yes at x=1

Answer 23

Thanks :)

Answer 24

yw