Proof Validation: of f(0)=0 given differentiable...

Question

Let $f$ be a differentiable function on an interval $A$ containing $0$, and assume $(x_n)$ is a sequence in $A$ with $(x_n)$ converging to $0$ and $x_n\neq0$ $\forall n\epsilon\mathbb{N}$.

Want to show: If $f(x_n)= 0$ $\forall n \epsilon \mathbb{N}$, show $f(0)=0$.

My proof so far:

By way of contradiction, assume $f(0) \neq 0$. Then, $f(0) = \alpha$, $\alpha\neq 0$.

Note, since $f$ is differentiable, $f$ is continuous.

Now, consider that $\forall\delta > 0$, $\exists N\epsilon\mathbb{N}$ such that $\forall n \geq N$, $|x_n-0|<\delta$. Take $n \geq N$. Continuity implies that $|f(x_n)-f(0)|<\epsilon$. Yet, $|f(x_n)-f(0)|=|0-\alpha|=|\alpha|$.

Then, the Archimedean Property states that $\exists \epsilon>0$ such that $0<\epsilon<|\alpha|$. Thus, $f$ is not continuous, a contradiction.

If this is all correct, I know I have the rest of my proof correct. Please let me know where (if anywhere) I make a mistake.

Thank you!

Answer 1

Suppose that $f(0)=a$ for $a \neq 0$. Then let $\epsilon=|\frac{a}{2}|$.

Clearly, there exists $\delta$ so that $|x|<\delta \implies |f(x)|<\epsilon$. But then there exists $N \in \mathbb{N}$ so that $|x_n|<\delta$. But then $|f(x_n)-f(0)|=|0-a|=|a|>\epsilon$, a contradiction.

Your idea was exactly right.