Find the values of a and b so that the following is true.

- anonymous

Find the values of a and b so that the following is true.

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

well, you can separate them,
then solve them

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

are you sure it tends to infinity?

- anonymous

Yep.

- anonymous

0.

- anonymous

oops.

- anonymous

why two variables?

- anonymous

if you differentiate the variable you get nothing from a, but b will still be present, you can get b right?

- anonymous

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

- anonymous

Yeah. I think you are right.

- anonymous

ahhh, so it tends to zero?

- anonymous

Yep :P . I typed wrong.

- anonymous

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

- anonymous

Hmm Let me try...

- anonymous

have you done it?

- anonymous

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

- anonymous

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

- anonymous

not 1, differentiate until the denominator is a constant

- anonymous

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

- anonymous

I got -4/3 .

- anonymous

for?

- anonymous

The limit.

- anonymous

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

- anonymous

how ?

- anonymous

To 0 :P .

- anonymous

lol!!! im losing it today...

- anonymous

but still, i would have thought it was zero

- anonymous

Waitt.

- anonymous

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

- anonymous

yep,

- anonymous

We can't solve anything then.

- anonymous

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

- anonymous

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

- anonymous

wait are you sure it should tend to 0?

- anonymous

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

- anonymous

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

- anonymous

not necessarily,
it can be infinity,
let assume it was infinity, then we can do the question with ease.
have you done analysis by any chance?

- anonymous

No.....

- anonymous

Here is the question.

##### 1 Attachment

- anonymous

okay, no problem,
assume it is infinity because i think you were right the first time round,
if it was infinity just plug in infinity to the differential you had computed

- anonymous

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

- anonymous

b would, it would be x^2,

- anonymous

one moment,

- anonymous

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

- anonymous

x^2 sorry not b^2.

- anonymous

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

- anonymous

Exactly. We can't work with that.

- anonymous

we could, if b = 0,

- anonymous

0/anynumber = 0

- anonymous

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

- anonymous

We don't know what b is.

- anonymous

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

- anonymous

But infinity- infinity is not 0.

- anonymous

yes it is

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

okay, you need to apply l'hpitals rule over and over again, i would recon,
but first you need to put all the equation under one common factor so
\[=\frac{\sin(2x) + ax^{3} + bx}{x^{3}}\]

- anonymous

http://answers.yahoo.com/question/index?qid=20110406112720AARpKn1
here is a better and easier way of doing it
the sint is taylor expansion i think,
the rest is pretty self explanatory

- anonymous

Lol. THanks :P .

Looking for something else?

Not the answer you are looking for? Search for more explanations.