Trouble with the following limit

1. Aug 16, 2005

devious_

I'm having trouble with the following limit:
$$\lim_{n \rightarrow \infty} 2^n \arcsin \frac{k}{2^n u_{n}} \text{, where \emph{k} is constant.}$$

I'm given that $\lim u_{n} = u$, where u is constant.

Apparently the book says the answer is $\frac{k}{u}$, but I can't figure out why.

2. Aug 16, 2005

lurflurf

Le Hopital's rules works well.
so does the substitution k/(2^n un)=sin(x)
if you have
lim x->0 sin(x)/x=1 as a known limit

3. Aug 16, 2005

arildno

Ok:
1. We have that $$arcsin(x)\approx{x},|x|<<1$$,
that is, when x is close to 0, arcsine is practically equal to x.
If you are unsure about it, remember that the SINE function sin(x) is practically equal to x when x is close to 0 (measured in radians, that is).
But:
Since arcsine is the inverse of sine we have:
$$arcsin(\sin(x))=x$$
by definition of the inverse.
For SMALL x's, we may replace sin(x) with x, and gets:
$$arcsin(x)\approx{x}$$
which is what we claimed..

2. Now, you should be able to do the rest..

Last edited: Aug 16, 2005
4. Aug 16, 2005

lurflurf

That works, but it is important to recall
Arcsin(x)=x+O(x^3)
lest one get confused when confronted with something like
$$\lim_{x\rightarrow 0}\frac{\sin^{-1}(x)-\sin(x)}{x^3}=\frac{1}{3}$$

5. Aug 17, 2005

arildno

Since it worked (hooray! it worked!!!), I didn't see any reason why I should load upon OP more than he needed.

And, if we are in need of the higher-order terms of arcsine, we can readily find them by inverting the power series of sine by the method of successive substitutions (or compute that Taylor series of arcsine otherwise, or look it up in a ready-made formula book etc.).

Last edited: Aug 17, 2005
6. Aug 17, 2005

devious_

Thanks guys. I was so determined that the book was wrong that I refused to think about this clearly. :)