Please explain use of Heine in proof of simple theorem

[twoflower]

Hi all,

I don't fully understand the proof of uniqueness of limit of function. Our teacher proved it using the Heine's theorem. Here it is:

Proof:
Let $\lim_{x \rightarrow a} f(x) = A$ and $\lim_{x \rightarrow a} f(x) = B$.

Let $\left{ x_{n} \right}$ satisfies: $\lim_{n \rightarrow \infty} x_{n} = a$. Then

$$\Longrightarrow^{Heine} \begin{array}{cc}\lim f\left( x_{n} \right) = A\\\lim f\left( x_{n} \right) = B\end{array}\right$$

Ok, I understand this, because according to Heine it's equivalent. But I don't get the next step:

$$\Longrightarrow^{\mbox{Uniqueness of limit of sequence}} A = B \Longrightarrow^{Heine 2} \mbox{Uniqueness of limit of function}$$

How can I use the uniqueness of limit of sequence here, when $f\left( x_{n} \right)$ is not a sequence, but a function?

Thank you for the explanation.

 

[HallsofIvy]

If lim(x->a) f(x)= A, then f(x_n) MUST also converge to A for any sequence x_n converging to a. If f converged to both A and B, then the sequence f(x_n) must converge to both A and B which, apparently, you have already proved is impossible.

Actually the proof of the uniqueness the limit of a sequence can be modified to give directly a proof of the uniqueness of the limit of a function.

If f converges to both A and B, take ε to be half the distance from A to B. Then show that x can't be within &epsilon of both A and B.

 

[twoflower]

...If f converged to both A and B, then the sequence f(x_n) must converge to both A and B which, apparently, you have already proved is impossible.

Well, that's what I don't understand. I think that $f\left(x_{n}\right)$ is function, not sequence. Sequence goes only over integers, whereas $f\left(x_{n}\right)$ doesn't...That's why I can't use the uniqueness of limit of sequence directly here I think...Of course there are other ways how to prove the uniqueness of limit of function, I just want to understand this one.

 

[arildno]

Why can't you construct the sequence:
$$a_{n}=f(x_{n})$$??