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Let $c\in \mathbb{R}$ and $f:\mathbb{R}\to \mathbb{R}$ is continuous at $c$. If for every positive $\delta$, there is a point $y\in (c-\delta, c+\delta)$ such that $f(y)=0$, show that $f(c)=0.$

I am unable to solve the problem. I don't know how to start the problem. I need a help.

Edit:

From the definition of continuity of $f(x)$ at $x=c$. For any $\epsilon > 0$, $\exists \delta > 0$:

$|f(x) - f(c)| < \epsilon \, \, \, \, \mbox{whenever} \, \, \, \, |x-c| < \delta$.

Let $x=y\in (c-\delta, c+\delta)$ st $f(y)=0$ then $|f(c)|<\epsilon$ whenever $ |x-c| < \delta$

then $f(c)=0$ (proved)

Whether the proof by $\epsilon-\delta$ method is correct?

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Take $\delta=1/n$, there exists $y_n\in (c-1/n,c+1/n)$ with $f(y_n)=0$, $lim_ny_n=c$ implies $f(c)=lim_nf(y_n)=0$ since $f$ is continue.

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  • $\begingroup$ Thanks a lot. Would you please comment whether my proof just edited is correct or not? $\endgroup$ – rama_ran Apr 9 '17 at 16:05
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Hint: $\delta_n := 1/n$. Then there exists $y_n$ such that $f(y_n) = 0.$ But $0 = \lim f(y_n) = f(\lim y_n)$ when $n\to \infty.$ But $\lim y_n = c.$

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