0

How can we use series representation in limits?

1) We can write $\sin x$ as $$\sin x=\sum\limits_{i=0}^\infty \frac{(-1)^ix^{2i+1}}{(2i+1)!}.$$

How can we write this?

For any given $\epsilon >0$ if we can find $N \geq 0$ such that $$\left|\sum\limits_{i=0}^N \frac{(-1)^ix^{2i+1}}{(2i+1)!}- \sin x\right| < \epsilon$$ than the equality $$\sin x=\lim\limits_{N \to \infty}\sum\limits_{i=0}^N\frac{(-1)^ix^{2i+1}}{(2i+1)!}=\sum\limits_{i=0}^\infty \frac{(-1)^ix^{2i+1}}{(2i+1)!}$$ will hold.

I am assuming we found this.

2) For $x\ne 0$ (we can show this in same manner) $$\frac{\sin x-x}{x^3}=\sum\limits_{i=1}^\infty \frac{(-1)^ix^{2(i-1)}}{(2i+1)!}$$ will hold.

3) Here, Our question is $$\lim\limits_{x\to0} \frac{\sin x-x}{x^3}.$$Using above series represenation we can write it as $$\lim\limits_{x\to0} \frac{\sin x-x}{x^3}=\lim\limits_{x\to0}\left(\lim\limits_{N \to \infty}\sum\limits_{i=1}^N\frac{(-1)^ix^{2(i-1)}}{(2i+1)!}\right). $$My question: How can we continue from here? If we can write $\lim\limits_{x\to0}\lim\limits_{N \to \infty}$ for $\lim\limits_{N\to\infty}\lim\limits_{x \to 0}$, how can we do that?

I want to give an counter-example for last part that limits not always commutative: $$\lim\limits_{x\to0}\left(\lim\limits_{y \to 0}\frac{x-y}{x+y}\right)=\lim\limits_{x\to0}1=1$$ and $$\lim\limits_{y\to0}\left(\lim\limits_{x \to 0}\frac{x-y}{x+y}\right)=\lim\limits_{x\to0}-1=-1.$$

I want to note that my question is not just for this particular example.

student forever
  • 1,266
  • Your series for $\sin; x$ is wrong: missing $(-1)^i$ – Robert Israel Apr 05 '16 at 15:27
  • Oh, thanks, I will edit now. – student forever Apr 05 '16 at 15:27
  • 2
    EXCELLENT question which shows you have the right attitude about mathematics. Whenever "we do something," QUESTION why we can do it. Perfect example - if you change the order of taking limits, WHY is that permitted (especially when you know it is NOT OK to do in general). You must be able to state specifically what it is, in your particular problem, that allows you to do something. In your example, it is not just the convergence of the series that allows you to change the order, but the UNIFORM convergence. –  Apr 05 '16 at 15:39
  • after correcting for the ${\left( { - 1} \right)^{,i} }$ what is now the connection between 1) and 2) ? cannot understand what is the problem – G Cab Apr 05 '16 at 15:43
  • First part is motivation for what I did and where I stuck. The question is after "My question" and it is about continuation the rest of calculation in right way. – student forever Apr 05 '16 at 15:46

2 Answers2

2

This has to do with special properties of power series; in particular the function defined by a power series is continuous on the interval where it converges:

Suppose $\sum a_nx^n$ converges for $|x|<R$. Fix $r\in(0,R)$, and say $r<\rho<R$. The series converges for $x=\rho$, and hence the terms are bounded for $x=\rho$; this shows that the series converges uniformly on $[-r,r]$, and hence the sum is continuous on $[-r,r]$.

David C. Ullrich
  • 89,985
0

Easiest way to do this is with $O$ notation. If $f(x) = \sum_{i=0}^\infty a_i x^i$, converging for $|x| < R$ where $R > 0$, then $f(x) = a_0 + a_1 x + \ldots + a_n x^n + O(x^{n+1})$ as $x \to 0$. So for example, $$\sin(x) = x - \dfrac{x^3}{6} + O(x^4)$$ which implies $$ \dfrac{\sin(x) - x}{x^3} = \dfrac{-x^3/6 + O(x^4)}{x^3} = -\dfrac{1}{6} + O(x) \to -\dfrac{1}{6} \ \text{as $x \to 0$}$$

Robert Israel
  • 448,999
  • This answer's seen as for calculation. But it wasn't the point what I didn't understand. Am I missing something in this answer? – student forever Apr 05 '16 at 15:40