Question

Can someone please explain to me why this limit doesn't exist?

Answer 1

\[\lim_{x \rightarrow \infty}(3x^2-1000x^2)\]

Answer 2

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/liminfsol1directory/LimInfSol1.html#SOLUTION 5
Can someone look at this page^ and look at solution 3, and tell me if this is right?

Answer 3

at first simplify inside the limit.

Answer 4

you get -997x^2, correct?

Answer 5

then, when x -> - infinity
there is essentially no limit

Answer 6

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}-997x^2=-\infty}\)

Answer 7

that means that limit does no exist.

Answer 8

because you can't get within (for example) 5 units from \(\large\color{slate}{\displaystyle-\infty}\)

Answer 9

I shared a link, do you mind checking it out? It says the limit goes to infinity.. Are they wrong?

Answer 10

for solution 3

Answer 11

when you have a limit = \(\large\color{slate}{\displaystyle\pm\infty}\) that means it does not exist

Answer 12

Ohh that makes so much more sense! So on an exam, if I have the limit goes to +/- infinity I should just right it 'does not exist'

Answer 13

Lets say
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}f(x)=100}\)
there is a particular limit, it is 100.
and you can get some particular distance away from 100. (which is not true about +- infinity)

Answer 14

yes, for limit to exist, it has to approach a particular value.

Answer 15

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}f(x)=100}\) Exist, it is 100

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty}f(x)=\pm \infty }\) Does Not Exist

Answer 16

well, on the exam, say that it approaches infinity and therefore diverges (or therefore dies not exist)

Answer 17

Limit of a function means that as you look at the graph, "what does the graph look like it will reach". If you have an infinity anywhere in your answer. that means the limit does not exist.

Answer 18

you want to do some other limits together, as examples ?

Answer 19

Sorry I lost the connection. Yes I would greatly appreciate that. This is my second time taking this course.

Answer 20

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~1^+}~\ln(x-1)}\)

Answer 21

how about this limit?

Answer 22

So ln(0)=DNE. Does it go to negative infinity?

Answer 23

it is not exactly ln(0), it is always a little greater than 0.

Answer 24

Can you tell me what is \(\large\color{slate}{\displaystyle\ln(0.1)}\) ?

Answer 25

what can you do to simplify it?

Answer 26

\(\large\color{slate}{\displaystyle \ln(0.1)=\ln(1/10)=\ln(10^{-1})=-\ln(10)}\)
correct?

Answer 27

Oh yes, I see

Answer 28

so, now lets take
\(\large\color{slate}{\displaystyle \ln(0.01)=}\)
please simplify it for me as I did with ln(0.1)

Answer 29

What if I had something like \[\lim_{x \rightarrow \infty}(\frac{ e^x-1 }{ e^x+1 })\]

Answer 30

you can differentiate on top and bottom

Answer 31

this method is known as L'Hospital's rule

Answer 32

Yes, I am familiar with that.

Answer 33

we are allowed to do so, because top and bottom the way they are now, BOHT, approach infinity.

Answer 34

Okay, I have to leave- I am in a club that is meeting at 8. But I greatly appreciate your help!

Answer 35

sure.....

Answer 36

I am glad when I help others, as I myself not good at math....
tnx and yw