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I am thinking about L'Hospital's rule but I am not sure how I would apply it...

well, you can separate them,
then solve them

Yeah.... But I have two variables then.

are you sure it tends to infinity?

Yep.

0.

oops.

why two variables?

\[\lim_{x \rightarrow 0} (\frac{ \sin(2x) }{ x^3 }+a+\frac{ b }{ x^2 })=0\]

Yeah. I think you are right.

ahhh, so it tends to zero?

Yep :P . I typed wrong.

no problem, you can separate them then do the limits,
you can spot it by inspection, very easily

Hmm Let me try...

have you done it?

I can't solve the limit of:
\[\lim_{x \rightarrow 0}\frac{ \sin(2x) }{ x^3 }\]

you can!!
differentiate 3 times until the denominator is 1,

not 1, differentiate until the denominator is a constant

sin(2)=2sin(x)cos(x)

I got -4/3 .

for?

The limit.

sin(2x)/x^3 as x tends to infinity?

how ?

To 0 :P .

lol!!! im losing it today...

but still, i would have thought it was zero

Waitt.

The Limit is infinity..... :( .

yep,

We can't solve anything then.

we could, i mean if one of the variables was infinity

But We can't do anything with infinity.... It's not a number.

wait are you sure it should tend to 0?

Well I would assume it should tend to a finite value.

i suspect the limit should tend to infinity then for it so satisfy the limit

No.....

Here is the question.

If we did that then a would be infinity and b would not exist.

b would, it would be x^2,

one moment,

but since b^2 tends to infinity the whole thing would be 0.

x^2 sorry not b^2.

which would be 0, at when b/x^2 as x ->0

Exactly. We can't work with that.

we could, if b = 0,

0/anynumber = 0

But it's not. B/x^2 is zero, not b.

We don't know what b is.

a could be -infinity,
then this would satisfy the limit

But infinity- infinity is not 0.

yes it is

No it's not. Infinity is not a number. It's a concept.

that is true, i moment, i am trying to multi task,
i jst remembered the proof that it isnt,

http://en.wikibooks.org/wiki/Calculus/Infinite_Limits/Infinity_is_not_a_number

Lol. THanks :P .