# Values of A and B in Limit

• Jul 22nd 2012, 03:35 PM
FrozenDragoon
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 $a$ and $b$ if
$\lim_{x\to\0} \frac{a+cos(bx)}{2x^2} = -9$

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
$a+cos(by)=0$
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,
$cos(by)=0$ or $cos(by)=-a$
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.
• Jul 22nd 2012, 03:50 PM
HallsofIvy
Re: Values of A and B in Limit
If this is a quiz are you allowed to ask for help?
• Jul 22nd 2012, 03:52 PM
FrozenDragoon
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.
• Jul 22nd 2012, 03:59 PM
FrozenDragoon
Re: Values of A and B in Limit
So I've kept working at it and I've determined that a can = -1

$a+cos(bx)=0$
$a+1=0$
$a=-1$

from there I plugged that value of a into the limit:
$\lim_{x\to\0} \frac{-1+cos(bx)}{2x^2} = -9$
and because that is a $\frac{0},{0}$ form, I applied L'Hospitals rule:
$\lim_{x\to\0} \frac{0+-sin(bx)}{4x} = -9$

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

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
• Jul 22nd 2012, 04:08 PM
Plato
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 $a=-1$.
• Jul 22nd 2012, 04:11 PM
Plato
Re: Values of A and B in Limit
Quote:

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

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

This is wrong: $b\ne 4$.
• Jul 22nd 2012, 04:16 PM
FrozenDragoon
Re: Values of A and B in Limit
Quote:

Originally Posted by Plato
This is wrong: $b\ne 4$.

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

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:
$\lim_{x\to\0} \frac{-b cos(bx)}{4} = -9$
in which $b=36$

The question implies there are many possible values (or maybe thats just the way I read it?), where might I go from here?
• Jul 22nd 2012, 04:22 PM
Plato
Re: Values of A and B in Limit
Quote:

Originally Posted by FrozenDragoon
Right, I know $b\neq 4$ but $b = 36$.

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:
$\lim_{x\to\0} \frac{-b cos(bx)}{4} = -9$
in which $b=36$

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

NO! $b^2=36$.
• Jul 22nd 2012, 08:01 PM
FrozenDragoon
Re: Values of A and B in Limit