# $f:\mathbb{R}\to \mathbb{R}$ is continuous at $c$. For $\delta>0$, there is $y\in (c-\delta, c+\delta)$ s.t $f(y)=0$, show that $f(c)=0.$

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?

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.
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 $\lim y_n = c.$