# What is lim(x->0)1/x=0?

1. Nov 12, 2006

### coldboyqn

I am doing something that I sure that I'm wrong, but I cannot realize the error. See as below:
$$\frac{1}{x}=x\times\frac{1}{(x^2)} ________$$1$$$$

Taylor Series of $$\frac{1}{x^2}$$:
$$\frac{1}{x^2}=\frac{1}{\alpha}+\sum_{k=1}^\infty g(k)(x-\alpha)^k$$

In which
k is from 1 to infinity,
$$g(k)=(-1)^k\times\frac{(k+1)!}{\alpha^{(k+2)}k!}\\\\ =\frac{(-1)^k\times(k+1)}{\alpha^{(k+2)}}$$

Substitute Taylor Series of 1/x^2 into (1), we obtain:
$$\frac{1}{x}=\frac{x}{\alpha}+\sum_{k=1}^\infty g(k)(x-\alpha)^k$$
So: $$\lim_{\substack{x\rightarrow 0}} \frac{1}{x}=\lim_{\substack{x\rightarrow 0}} (\frac{x}{\alpha}+\sum_{k=1}^\infty x\times g(k)(x-\alpha)^k)=0 (!?!?!?)$$
Can any one show me, please?

Last edited: Nov 12, 2006
2. Nov 12, 2006

### StatusX

You need to check the radius of convergence of the taylor series. Note if you don't multiply by x, you get the limit as x->0 of 1/x^2 to be 1/a^2, where a is arbitrary. (I'm assuming alpha=a^2)

3. Nov 12, 2006

### coldboyqn

I see that if using the Taylor series above to determine the value of $$\lim_{\substack{x\rightarrow 0}}\frac{1}{x^2}$$ we will obtain infinity, which is according?

4. Nov 13, 2006

### StatusX

You're right. Sorry, I should have read your question more carefully. The problem is that the last limit isn't 0. You can't just plug in 0 to get the limit, as the function isn't continuous at x=0. More careful calculation should show that limit diverges as well.

5. Nov 13, 2006

### JasonRox

Wouldn't it be best to draw a graph? Especially when it's possible.

6. Nov 14, 2006

### coldboyqn

Surely the last limit isn't 0. But my problem is that the strange result I obtain when
I treat $$\frac{1}{x}$$ as $$x\times TaylorSeries\_of(\frac{1}{x^2})$$.
I wonder where is my error when I calculate the limit by this method!

7. Nov 14, 2006

### StatusX

It isn't hard to check the series converges precisely in (0,2a), and in this region it converges to 1/x^2. If you multiply it termwise by x, you get a series that converges in [0,2a). But there's no reason to expect that series evaluated at x=0 to give you the same thing as (1/x^2)*x evaluated at 0, since the taylor series did not converge at x=0. You can prove the limit diverges explicitly, and the easiest way to do this is just to prove the series does converge to 1/x in (0,2a).

8. Nov 15, 2006

### coldboyqn

Oh, yes, I see. As the Taylor Series of $$\frac{1}{x^2}$$ diverges at 0, I cannot simply multiply it with x to evaluate 1/x at 0, right? And it is unreasonable to multiply an expression that diverges (to $$\infty$$) with a variable that come to zero and conclude that the multiplied expression come to zero, right?
Thanks for explanation, I understand now.