Question

how do i determine the infinite limit of: lim x->3+ f(x)=ln(x^2-9)

Answer 1

its limit does not exist

Answer 2

hes talking from the positive direction i think

Answer 3

ln(0) = negative infinity

Answer 4

can you explain why it does not exist?

Answer 5

the x->3+ means the right side correct?

Answer 6

yes

Answer 7

how did you get ln(0)?

Answer 8

im thinking something like

if x>3 since its approaching from the positive side

and x^2-9

3.000001^2-9 = some positive number infinitely close to 0.


if we were coming from the negative side x<3

2.999999^2-9=some negative number infinitely close to 0.

Answer 9

however the vertical asymptotes are at -3 and 3

Answer 10

because you factored it out right

Answer 11

so x cannot equal 3 or -3

Answer 12

correct, but approaching from the positive side it goes to negative infinity, im trying to figure out how to word this right..

Answer 13

ok wait so i was told that if lim x->a- f(x) does not equal to lim x->a+ f(x) then the limit does not exist
in other words if the left side and right side are not the same then the limit does not exist

Answer 14

true

Answer 15

i think the ln is just throwing me off in this one

Answer 16

x^2+9 has to be greater then or equal to 0

ln(x)=y
e^y=x

x has to be greater then 0

Answer 17

x^2+9 nvm cant be equal to

Answer 18

but it can be infintely close (approaching 3)

Answer 19

but it never reaches it right?

Answer 20

no, but it could be 0.000......1

if you tried approaching from the negative side you would have a negative number close to 0. but it has to be greater then 0, so it can only approach from the positive side

Answer 21

i had a quick question you do not plug in the 3 into the equation right? because i know there are some limits that you do to try and find what type it is

Answer 22

so would the answer be +infinity or no?

Answer 23

i never plugged in 3, just thinking conceptually.

try graphing it, you will see the asymptotes and it goes down.

with ln(x), x cannot be less then 0

instead were saying ln(x^2-9)

when x approaches 3 from the positive side its sum number greater then 3 but really close to 3. like 3.00000000000001^2-9

3.0000000001^2-9=some positive number really close to 0

for ln(x), as x approaches 0 from the positive side it goes to negative infinity

Answer 24

oh ok so you know the limit laws? those are used for other types of limits?

Answer 25

im taking the same class as you calc 1 right?

sept im like a week ahead i think...so im no expert on this

Answer 26

yea its called math 2a(calc) here i have a quiz tomorrow in discussion

Answer 27

id have to work on this for 40 minutes to give a clearer answer

Answer 28

what book are you using for your class? wonder if its the same

Answer 29

its the schools custom book. same material probably

Answer 30

oh we are using the Stewart's Early Transcendentals 7th ed

Answer 31

i dont think there are any limit laws in play here since the limit DNE.

it only exists when approaching from the positive side.

Answer 32

we are on section 2.6 but are getting quizzed on 2.2. and 2.3

Answer 33

oh ok

Answer 34

like theres some where you have to calculate whether the limit is infinite or not and then it has to do with the numerator and denominator
idk if that applies here

Answer 35

\[(3^-)^2 < 9 \]

\[(3^+)^2 > 9 \]

dont think theres any num/denom stuff here

Answer 36

@SithsAndGiggles

Answer 37

ok so are possible answer choices infinity, -infinity, or DNE ? for these type of probs

Answer 38

this one is DNE right?

Answer 39

if the question is\[\lim_{x \rightarrow 3^{+}} \ln(x^2-9)=-oo\]
\[\lim_{x \rightarrow 3^{-}} \ln(x^2-9)=DNE\]

\[\lim_{x \rightarrow 3} \ln(x^2-9)=DNE\]

Answer 40

and thats because the domain is

(-oo,-3]U[3,oo)

Answer 41

ohhhh ok now i see what you're saying

Answer 42

so it doesnt exist

Answer 43

so since it asked for the right side x->3+ then as it get closer to 0 then it is going in the direction of -infinity? or am i misunderstanding still?

Answer 44

yes

Answer 45

the limit DNE, but the limit from the positive side is at negative infinity

Answer 46

since the section your on is limits at infinity im assuming the answer is gonna be either negative or positive infinity.

Answer 47

ok so why isn't the other on x->3- at positive infinity?

Answer 48

because of the domain restriction

Answer 49

the function would go to positive infinity as x goes to positive or negative infinity

Answer 50

do you know what the graph looks like?

Answer 51

https://www.desmos.com/calculator try this

Answer 52

there are two V.A right?

Answer 53

ya

Answer 54

oh cool thanks

Answer 55

thatzz what i said loll

Answer 56

ok its like going downward