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Prove that if $f$ is odd then $f(x) = 0$ for $x = \pi$

Prove that if $f$ is odd then $f(x) = 0$ for $x = \pi$
Suppose $f(x) = g(x)$ and $g(x)$ is even. Then $f(x) = -f(-x)$ but $-\pi = \pi$ so we have a contradiction.
Is this proof correct?

A:

There is a much simpler proof. Let $f(x)$ be even, that is,
$$
f(x)=f(-x).
$$
Then
$$
f(\pi)=f(\pi+\pi+\cdots+\pi)=f(0)=0.
$$

A:

You can prove that if $f(x)$ is odd then $f(x)=0$ for $x=\pi$ in many ways, not only by using continuity. Here is a standard proof that’s easy to understand.
For $\alpha\in\mathbb R$, let $x_\alpha:=\pi+\alpha$. Then $x_\alpha=\alpha\pi$, so $x_\alpha\to\pi$ and $f(x_\alpha)\to f(\pi)$ as $\alpha\to0$. It follows that
$$
f(\pi) = \lim_{\alpha\to0} f(x_\alpha) = \lim_{\alpha\to0}\lim_{\beta\to0}f(x_\alpha+\beta) = \lim_{\beta\

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