1. Jul 17, 2015

### GiuseppeR7

Hi guys...i'm a little naive...i encountered the limit of this function:

Sin(x^-1) x

as the x goes to infinity...in order to study it i know that i have to find the Taylor series about the function Sin(t) centered in 0 having defined t=(x^-1)...something called asymptotical expansion of Sin(x^-1). The fact is that i have not found this technique or the theory behind this so called "asymptotical expansion" in any book! So i was asking of somebody can help me about this with some explanation or some material! thank you!

2. Jul 17, 2015

### RUber

Asymptotic expansion is a process where you expand a function about its limit.
In $x \sin \frac1x$ with a large x, you have one large term and one small term, so asymptotic expansion is one good way to understand the behavior of the function for large x.
A quick search pulls up plenty of resources. One that looks reasonably explanatory is http://www.math.ubc.ca/~feldman/m321/asymptotic.pdf.

3. Jul 17, 2015

### HallsofIvy

You should know that $\lim_{\theta\to 0} \frac{sin(\theta)}{\theta}= 0$. That is the same as saying that, for small $\theta$, $sin(\theta)$ is approximately equal to $\theta$ and the approximation gets better the smaller $\theta$ is. As x goes to infinity, $\frac{1}{x}$ goes to 0 so $\lim_{x\to \infty} x sin(1/x)= \lim_{\theta\to 0}\frac{sin(\theta)}{\theta}= 1$.

4. Jul 17, 2015

### GiuseppeR7

Ok, thanks for the reply...for example...how can i find the asymptotic series for Ln(1/Sqrt(1 + x)) ?

Last edited: Jul 17, 2015
5. Jul 17, 2015

### RUber

I would go to wolframalpha.com and type in "series ln(1/(sqrt(1+x)))".