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\[\lim_{x \rightarrow \infty} [\sqrt[3]{n ^{2}}/n+1]\]
Is it Zero?
@Ishaan94 Are you asking me? O.o...idk....i think it is inifinite/infinite =udefined..IDK :S

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Other answers:

\[\large\frac{\sqrt[3]{\frac{n^2}{n^3}}}{1 + \frac1n}\]
Huh???? I didn't get it AT ALL Ishaan...exaplin it please
explain*
Divide and Multiply by n.
Ishaan can u draw it step by step please
I thought apoorv was doing it.
|dw:1334846528832:dw|
\[\Large \frac{\frac{\sqrt[3]{n^2}}n}{\frac{1+n}n}\]
My tab "Aw, Snapped" me :'( (lost my labor *sniff sniff*) ^^what Ishaan said.
@Ishaan94 would I sound way too dumb if i told you that I STILL don't get it O.o
No.
Am I only one who this that the limit is \[ \lim_{x \rightarrow \infty} \frac{\sqrt[3]{n ^{2}}}{n}+1 \]
HOW? then what were we writing? And why did Angela not correct us?
@FoolForMath it is not like that..it is |dw:1334846830006:dw|
|dw:1334846790537:dw|Why can't I draw better?!?!?!
Your latex is misleading.
seems like rockstart's gonna rock us now.
lol
|dw:1334846742289:dw|
|dw:1334846901905:dw|
Yes. Yes. Do it as apoorv says.
@apoorvk: That's not true.
Yayy yayy.. @rockstar111 baba's here :p
If \(n \to \infty \) then \(\frac 1n \to 0\).
I am wrong? -\(o.O)/_
thats.what.i.did.
comparing two infinities is blasphemy alright :/ But then, how do I express myself?
Okay it's specious.
No need to write in terms of infinity, just use the limits and show the answer.
Like the way FoolForMath did.
I am sure @angela210793 has gone crazy by now.
|dw:1334847232658:dw|
I am not saying that your way is incorrect, I think it might be accepted in some places.
but that is how I would do this one.
to be honest, I wouldn't blame her from spraining a brain muscle after all this
I don't mean to confuse.
rigor is important as well, so I get your drift :-)
She left :/ I won't be a good professor/teacher.
I didn't leave....mozilla froze.....and thank you all...:))) (Idk whom to give a medal now :S)

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