need help with limit graph and q's

- anonymous

need help with limit graph and q's

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- anonymous

can you see it

##### 1 Attachment

- anonymous

a) = -1
b = 2

- anonymous

hmmm I don't think that is right a or b

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## More answers

- anonymous

what do you think it is

- anonymous

limit is asking what is the y value as x gets closer to -3 ...a) is asking what is y becoming as x turns into three from the left side. That arrow is definitely not pointing to -1. b) is asking the same thing from the right side that arrow is not pointing at 2.

- anonymous

x turns to negative 3

- anonymous

so it should be -1 right?

- anonymous

-3 looks like a vertical asymptote to me...which x=-3 are you looking at? maybe the graph you posted and the one im seeing are different?

- anonymous

oooh my mistake

- anonymous

so - inf

- anonymous

for a)

- anonymous

3 for b?

- anonymous

yes but depending on how your teacher/book are looking at it they may say that the limit is undefined too
if you start at x = 0 and move left towards x=-3 what happens to the line with the arrow?

- anonymous

yeah undefined also

- anonymous

=) that one is heading to +oo

- anonymous

so I know you have c) nailed

- anonymous

and d) for that matter

- anonymous

c = DNE

- anonymous

and neg inf

- anonymous

d = 3

- anonymous

oh so close....c) is saying if you approach from either direction...so it is + and - infinity depending which side of the line you are on....so undefined probably best.
for d) what is y when x=-3?

- anonymous

d = 3 right

- anonymous

no =) the arrowy lines don't point at 3 they point at?
-oo and +oo.....so .......undefined the lines dont cross x=-3 at all

- anonymous

alright the next few are tough

- anonymous

is it all .5?

- anonymous

e-h

- anonymous

uh oh I thought there were getting easier...
all good

- anonymous

i - L have a trick to them

- anonymous

wait so e- h = .5 right?

- anonymous

indeed you are correct

- anonymous

cool

- anonymous

i = 1

- anonymous

f = 1

- anonymous

it is always easier when the limits exist
i=1 is correct

- anonymous

well j anyway =)

- anonymous

right haha

- anonymous

k also 1

- anonymous

you on a roll now

- anonymous

j = dne

- anonymous

woot

- amistre64

j should exist; its a onesided limit

- anonymous

nah it cant its asking for the value at f(0)

- anonymous

oh right yeah

- amistre64

j. lim{x ->0+} = ?? its goes to 1

- anonymous

lets move onto m

- anonymous

aaak i didnt pay attention to you going back to j I read that as an L

- anonymous

yep we already got that

- anonymous

yep me 2

- anonymous

okies I am at M

- anonymous

m = 4?

- anonymous

f = 4

- anonymous

sorry n = 4

- amistre64

m=5 if the graph is right

- anonymous

m != 4 keep in mind the axis are in a funny spot

- anonymous

yeah

- anonymous

x=1 looks a lot like that vertical asymptote at -3 to me

- anonymous

m-p = 4

- anonymous

I think you are looking at the wrong x=1

- anonymous

oh hi mistre!

- anonymous

yep

- anonymous

m= - inf

- anonymous

I think so too

- anonymous

n = - inf

- anonymous

now we back on track

- anonymous

o = - inf

- anonymous

o. oh oh yeah

- anonymous

and p = dne

- anonymous

=) you got it now

- anonymous

only q 2 p 2 go

- anonymous

q = - .5

- anonymous

or maybe that should be a t who built this chart anyway?

- anonymous

r = -.5

- anonymous

yeah really weird graph

- anonymous

I buy q=-.5 and r too for that matter

- anonymous

s = -.5

- anonymous

yay

- anonymous

q as well -.5

- anonymous

t) i mean

- anonymous

ding you have reached the next level....hehe alphabetically challenged homework

- anonymous

new graph hold on second

- anonymous

do you see it

##### 1 Attachment

- anonymous

limx→2+f(x) =5≠limx→2−f(x)=2

- anonymous

let me print it out standby

- anonymous

we need 4 answers

- anonymous

and im playing world of warcraft too....semester ended monday for me =)

- anonymous

haha

- anonymous

are you in hs or college?

- anonymous

college...again ....

- anonymous

nice pass time hit the ladies up instead ;)

- anonymous

but what would my wife think?

- anonymous

ah never mind then

- anonymous

f(2) exists (infact f(2) = 4

- anonymous

we need 4 answers

- anonymous

f(x) is continuous at 2 i think jump continutity

- anonymous

so far
a)f(x) is continuous at 2 i think jump continutity
b) f(2) exists (infact f(2) = 4
c) f(2) = limit x->2 F(x)

- anonymous

jump continuity is the opposite of continuous though

- anonymous

so is it continuous?

- anonymous

I'm a little confused by the notation for the equations...

- anonymous

check this file out

##### 1 Attachment

- anonymous

oh i get it as x ->2+ f(x)=5
I was being dense

- anonymous

thats what the teacher did on her practice pdf

- anonymous

so i am assuming we need 4 answers

- anonymous

ok so from the right side of the graph we have a line from +oo to 2 running along y=5

- anonymous

alright you mean in limit notation

- anonymous

I'm just graphing it atm....but yes the notation was throwing me off for a moment I mentioned the other night that math was hard enough and typing math made it math^2
from the left side of the graph a horizontal line from -oo, 2 running along y=2 is that what your graph looks like?

- anonymous

no no dont go by the 2nd attachment that was just an example we are still working on the first

- anonymous

this one

##### 1 Attachment

- anonymous

This one limx→2+f(x) =5≠limx→2−f(x)=2

- anonymous

yeah

- anonymous

i just pasted the other graph to show you a comparison of what the teacher expects i assume on this one

- anonymous

since the left and right side limits at 2 are not equal....it is not continuous...or if you are a more graphy person: you can't draw that graph without lifting your pencil

- anonymous

cool i will figure out the rest enjoy warcraft

- anonymous

thank you for all your help!

- anonymous

okay you enjoy limits...bwah haha
and the book calculus for dummies really helped me with limits this semester

- anonymous

cheers man.. take it ez

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