Question

Can someone help me on this problem..? It is attached.

Answer 1

if you show me how to find it, I'll do it on my own. I'm not expecting answers for this huge problem.

Answer 2

Just follow the graph lines. So like:
\[\lim_{x \rightarrow 2^{-}}\]
You would simply trace your finger along the curve starting on the left hand side until you get to 2. Whatever y-value it's approaching at the point x=2, that's your answer :)

Answer 3

as g approaches 0 from the left, and also from the right, the function is approaching some number a little larger than 3, the \(y\) value

Answer 4

So as x approaches 2- ... its 3?

Answer 5

as @nj1202 said.
and if you are approaching 2 from the right, that is for \(\lim_{x\to 0^+}\) trace your finger on the curve going from right to left and see what \(y\) value you are approaching

Answer 6

yes

Answer 7

2 from the right= 1?

Answer 8

yes
only tricky one is
\[\lim_{x\to 2}\]

Answer 9

The limit doesn't exist?

Answer 10

and since the above answers are not the same (that is \(3\neq 2\)) then that limit does not exist
yes

Answer 11

So it's no number, its just the the DNE.

Answer 12

limit*

Answer 13

yes

Answer 14

what about g(2)? that's different.

Answer 15

Just a helpful hint for when these are loaded into a test. NEVER!! EVER!! write that the limit DNE for a one-sided limit. Only when it's approaching from both sides can you say DNE.

Answer 16

g(2) means x=2, so just find that point.

Answer 17

which point do I go to though?

Answer 18

for x=2 theres two?

Answer 19

Yes, but notice how one is shaded and the other is a circle. The circle means that the point is not defined at that x value. The shaded means that it is. It's called a jump in a function. g(2) = 3

Answer 20

why is it 3 (shaded) and not 1(unshaded)?

Answer 21

Because the shading refers to where the point is actually defined. Think of when you draw a graph. If you were to simply plot a point on a cartesian plane, what would you do? You'd shade a small circle onto the graph and write the coordinates. The reason it's not 1 is because it's unshaded. Unshaded means that the x-value is not included in the domain of the function.

Answer 22

ohh.. okay. So for any number to be in g(?)... it has to be included. its never an unshaded point.

Answer 23

Also, is as the limit approaches 4 g(t)=4?

Answer 24

Actually i can't see the picture, is that another unshaded circle? at x=4?

Answer 25

and its continuous because of the jump continuity?

Answer 26

Well, usually we just say jump discontinuity if it's not continuous. So it's just continuous. The thing is I missed x=4 and there looks like an unshaded circle, so it's undefined, and not continuous like i thought it was xD

Answer 27

when I look at x=4 and follow it up to y there is an unshaded circle at (4,4)

Answer 28

Ok, I couldn't tell in the photo. So yea, the limit still exists though, because regardless of if you approach x=4 from the left or from the right, you approach y=4, so since they are the same, it exists even at a point of being undefined. Tricky question :)

Answer 29

so... as the limit approaches 4, g(t) is undefined.

Answer 30

im a lil confused now haha..

Answer 31

haha i'm sorry xD i'll get this through to you! :D

okay, so the point g(4) is undefined, because it is unshaded.

however, look at both one-sided limits of x=4.
x=4 from the left, you approach y=4
x=4 from the right, you approach y=4

since the two one-sided limits are the same, the limit exists

Answer 32

okay, so g(4) is undefined. However as x approaches 4- its 4 and 4+ its 4. the limit of g(4) exists making g(4)=4 ? is that right?

Answer 33

Well, no g(4) does not = 4, the limit as x-->4 is 4.

Answer 34

g(4) is undefined. then But, As x approaches the limit of 4, g(x)=4?

Answer 35

Yup. That's it!

Answer 36

holy crap. time for a nap. haha

Answer 37

haha wish i could say the same, i got calc III at 5pm xD

Answer 38

aww, :/ one more question and i'll try to understand faster.. lol!

Answer 39

no problem. my walk is 5 minutes. what's up? :D

Answer 40

The question is wanting me to find the limit as x approaches 0 from both sides... but when I go to x=0 the y is only a little bit above the three mark.. and no shaded circle or unshaded?

Answer 41

Yea, i saw that. You would just put the limit is approaching a little more than 3. I don't know how you'd write that haha xD It's like 3.1 or 3.2 i guess xD but the same rule should apply to it.

Answer 42

So for 0- would it be g(x)=3.1? like that...?

Answer 43

Yea that shoud be fine, that is what it's approaching.

Answer 44

I don't know why the people would be mean and ask for something just randomly off a line, but oh well xD

Answer 45

and is it the same for 0+?

Answer 46

haha yeah well, my professor... uhm, I can't understand her. (Chinese) :/ so I have to learn everything on my own....

Answer 47

yup exactly the same! oo, accents are hard sometimes haha xD

Answer 48

and as x approaches 0 g(t)= 3.1 as well and the limit exists?

Answer 49

It certainly does! :D

Answer 50

and yes, it sucks!

Answer 51

haha my only foreign teacher this semester is my EM theory xD he's german, so it's quite gutteral xD

Answer 52

my goodness, thank you so much! I wish I could like take you to lunch or something instead of giving you a best response lol.. I have quite a few:/ I guess that's what you get for being in engineering.

Answer 53

Haha I'm astronomy/astrophysics so i'm in the math business too. I'm usually one quite a bit, just shoot me a question :D

Answer 54

I was going to do astrophysics! but decided on Aerospace Engineering (Astronautics). and okay will do!

Answer 55

Sweet deal buddy :D Space geeks will rule the world! Don't let anyone tell ya otherwise haha xD

Answer 56

hahaha agreed! talk to you later :) thanks so much again!