Values of A and B in Limit

This is the last problem on a take home quiz I was given. I keep staring at it, but I'm not really even sure how to start.

Determine all possible values of $\displaystyle a$ and $\displaystyle b$ if

$\displaystyle \lim_{x\to\0} \frac{a+cos(bx)}{2x^2} = -9$

I've done some searching, and found 2 instances of a similar problem:

Yahoo! Canada Answers - Find Numbers a and b for a limit?

Find numbers a and b? - Yahoo! Answers

In both of those problems, the denominator in the limit is only X, so they state that as x->0, the numberator must be equal to zero as well.

Ok, so

$\displaystyle a+cos(by)=0$

in both the examples that I have found, the person working it out would put in a 0 for a at this point. Ok, but If I do that,

$\displaystyle cos(by)=0$ or $\displaystyle cos(by)=-a$

That cannot be broken up anymore. In the examples, after finding b using this method, they use it to find other values. But without a way to isolate b I'm stuck.

Thanks in advance for any help.

Re: Values of A and B in Limit

If this is a **quiz** are you allowed to ask for help?

Re: Values of A and B in Limit

Yes, he encourages us to ask for help, and even allows groups of students working together to turn in a single set of answers/work for credit.

Re: Values of A and B in Limit

So I've kept working at it and I've determined that a can = -1

$\displaystyle a+cos(bx)=0$

$\displaystyle a+1=0$

$\displaystyle a=-1$

from there I plugged that value of a into the limit:

$\displaystyle \lim_{x\to\0} \frac{-1+cos(bx)}{2x^2} = -9$

and because that is a $\displaystyle \frac{0},{0}$ form, I applied L'Hospitals rule:

$\displaystyle \lim_{x\to\0} \frac{0+-sin(bx)}{4x} = -9$

I know that if $\displaystyle b=4$ then the limit equals 1, but I'm not sure if the rule applies to multiples (ie: $\displaystyle \lim_{x\to\0} \frac{sin((8)x)}{4x} = 2$?).

If it does hold true then b would be equal to 36.

Edit: just tried it and it does seem to hold true (I just dot remember seeing it anywhere).

So, A = -1 B = 36

Re: Values of A and B in Limit

Quote:

Originally Posted by

**FrozenDragoon** Yes, he encourages us to ask for help, and even allows groups of students working together to turn in a single set of answers/work for credit.

Apply l'Hopital"s rule twice. But to do so we must have $\displaystyle a=-1$.

Re: Values of A and B in Limit

Quote:

Originally Posted by

**FrozenDragoon** I know that if $\displaystyle b=4$ then the limit equals 1, but I'm not sure if the rule applies to multiples (ie: $\displaystyle \lim_{x\to\0} \frac{sin((8)x)}{4x} = 2$?).

If it does hold true then b would be equal to 36.

This is wrong: $\displaystyle b\ne 4$.

Re: Values of A and B in Limit

Quote:

Originally Posted by

**Plato** This is wrong: $\displaystyle b\ne 4$.

Right, I know $\displaystyle b\neq 4$ but $\displaystyle b = 36$.

I found this two ways, the first by pluging in 36 before applying L'Hospitals rule a second time

Wolfram Alpha

and second, after using the rule a second time I get:

$\displaystyle \lim_{x\to\0} \frac{-b cos(bx)}{4} = -9$

in which $\displaystyle b=36$

The question implies there are many possible values (or maybe thats just the way I read it?), where might I go from here?

Re: Values of A and B in Limit

Quote:

Originally Posted by

**FrozenDragoon** Right, I know $\displaystyle b\neq 4$ but $\displaystyle b = 36$.

I found this two ways, the first by pluging in 36 before applying L'Hospitals rule a second time

Wolfram Alpha
and second, after using the rule a second time I get:

$\displaystyle \lim_{x\to\0} \frac{-b cos(bx)}{4} = -9$

in which $\displaystyle b=36$

The question implies there are many possible values (or maybe thats just the way I read it?), where might I go from here?

NO! $\displaystyle b^2=36$.

Re: Values of A and B in Limit

Oh... Your right XD thank you. I made the mistake of not checking my answers

But... why is it 6? and not 36?

Does it have to do with the fact that 36 was obtained through deriving the function?