Question

State the value of this
quantity, if it exists. If it does not exist, explain why:
h(-3)

Answer 1

it doesn't exist unless we know what \(h(x)\) is

Answer 2

I'll upload the accompanying graph

Answer 3

Does that change anything?

Answer 4

By "know what h(x) is" do you mean knowing the limit?

Answer 5

There is a hole at x = -3 on the graph, so it looks like h(-3) doesn't exist and it's not defined.

Answer 6

That makes sense! What would the case be with h(0)?

Answer 7

you tell me

Answer 8

Well I don't know... if we're coming from both sides, one has a hole and one doesn't

Answer 9

but do you see how there's a closed circle at (0,1)

that means that h(0) is defined

Answer 10

the limit is NOT defined at x = 0 because the left and right hand limits are different as x--> 0

but the limit is defined at x = -3

Answer 11

Ok thank you

Answer 12

yw

Answer 13

Wait... you reasoned that h(-3) cannot exist due to the presence of a hole, bu then by that logic, how would \[\lim_{x \rightarrow 0^+} h(x)\] exist, since it too encounters a hole

Answer 14

h(-3) itself doesn't exist, but the limiting value as x ---> -3 does exist

Answer 15

h(0) exists because h(0) = 1

Answer 16

but as x ---> 0, the limit is NOT defined (the two sides need to meet up at the same y value for the limit to exist)

Answer 17

limits can exist if you approach a hole...you're not actually arriving at the hole, just getting closer and closer

Answer 18

Oh, that makes sense! So h(-3) doesn't exist do to the fact that it reaches (-3) and encounters a hole, while limx→0+h(x) can exist because it's only approaching (that it, never reaching) the 0

Answer 19

Thank you again!!

Answer 20

h(-3) also doesn't exist because plugging in x = -3 doesn't produce any y value or output, so that's why that hole exists and there are no other points above or below that hole (that we can see at least), so this adds more evidence that h(-3) doesn't exist at all

Answer 21

and yes the right hand limit exists because you can get closer and closer to 0 from the right (this works because the function is continuous to the right of x = 0, but not actually continuous at x = 0)

Answer 22

Superb!! So just one more question... you said above that h(0)= 1, but how is that so if there's two dots... do we just assume the one that is a hole to not exist, so we go with the positive one (1)?

Answer 23

the function exists at the closed circle

the open circle denotes that something should exist there, but it's like the point has be ripped out and moved elsewhere

Answer 24

for instance, say we had this continuous graph

Answer 25

the point (5,2) exists and it is defined, so if this graph is of f(x), then f(5) exists and it is defined