Question

I will post the question as an image. It is on limits and continuity.

Answer 1

The question is: Find a value "a" where neither of the one-sided limits exists, but f(a) exists. What do we write to describe the behavior near a?

Answer 2

What did you get for 'a'

Answer 3

I got that f(2) exists

Answer 4

But I dont understand the 2nd part to describe teh behavior near a.

Answer 5

But is f(2) correct?

Answer 6

no, 'a' isn't 2

notice how the LHL and the RHL both exist for x = 2

Answer 7

But its an open circle for the LHL of f(2)

Answer 8

Im confused. What do I look for when looking for teh a?

Answer 9

where are there infinite limits?

Answer 10

at f(4) ?

Answer 11

good, so the LHL and the RHL don't exist there

Answer 12

but f(4) does exist, so a = 4 is the answer

Answer 13

Oh I see. But now how do I describe the behavior near a? Thats teh 2nd part of teh question.

Answer 14

as x gets closer and closer to a = 4, f(x) gets very very large

Answer 15

so it's effectively approaching infinity, which means that the LHL and the RHL don't exist (which means that the limit itself doesn't exist either)

Answer 16

Is tehre a way to write that using notations.

Answer 17

you mean using delta and epsilon?

Answer 18

I thought u can use like limx->a etx...like that

Answer 19

oh you can do that as well

\[\Large \lim_{x\to4^{-}}f(x) = \infty \]

\[\Large \lim_{x\to4^{+}}f(x) = \infty \]

\[\Large \lim_{x\to4}f(x) = \infty \]

Answer 20

Oh I see. There are other parts of this q related to teh graph. Can youtell me if my answers are corect?

Answer 21

sure

Answer 22

I'm just going to upload the pic of teh q's but I dont need help with all teh questions.

Answer 23

I just dont feel like typing...lol

Answer 24

For d) I'm a bit confused as to what that limit means.

Answer 25

Is it all the places where there are closed circles?

Answer 26

it's anywhere on the graph where if you approach that point from both sides, you'll end up at f(a)

Answer 27

So like at a= -2 , 0 ?

Answer 28

no a = -2 is a bad example

Answer 29

if you approach -2 from both sides on the graph, you'll end up at the point (-2, -2)

BUT

f(-2) is NOT -2, it's actually f(-2) = 2

Answer 30

Oh I see. i think that answers e) right. But is a=0 correct?

Answer 31

B/c all the other points have both open adn closed circles at values of a when i pick them

Answer 32

e) for a = -2, the limit exists, but f(a) is not equal to the limit

Answer 33

And for d) was 0 correct? It says there are many values of a but im not sure.

Answer 34

d) a = 0 is correct

Answer 35

I think 0 is also the answet to f) right? B/C the LHL and RHL is 0 at f(0) but f(0) does not exist

Answer 36

i think there's a hole at x = 0, is that right?

Answer 37

yes terhe is

Answer 38

so yes, the limit exists but f(0) doesn't

Answer 39

Sorry Im the same person. its just that i cant log back into this session so I just had to make a new account :/ and lastly for g) is the asnwer a=2

Answer 40

oh weird...

Answer 41

yes, for g), the answer is 2

Answer 42

And h is that it means its continius when: lim x->a f(x) = f(a)

Answer 43

THANK YOU SOOO MUCH !! :DDDD YOU HAVE BEEN A TREMENDOUS HELP

Answer 44

exactly, f(x) is continuous at x = a if

1) f(a) is defined
2) lim x -> a f(x) is defined and exists
3) lim x -> a f(x) = f(a)

Answer 45

Thanks for the thorough answer. I really appreaciate your help and patience. Sorry I'm just so slow at calc :/

Answer 46

you're fine, glad to be of help

Answer 47

:) Yes you were amazing help. Good night.

Answer 48

good night