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Continous function with infinitely many zeroes

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Let's say I have a function $f(x)$, that is continous everywhere on the real line. Suppose I have a sequence of real numbers $a_k$ such that $a_0$ = $C$ and as $k$ tends to infinity, $a_k$ tends to a real finite number $L$. Thus $a_k$ is Cauchy, and around $L$ the sequence would get arbitrarily close. Now, if $f(a_k)=0$ for all $k$, does it follow that $f(x)=0$ for all $x$? I've tried thinking of counter examples, and one that came to mind was $sin(\frac{1}{x^2})$ around $x=0$. However at that point that function is no longer continous, so would it follow then that a function that satisfies those conditions is $0$ everywhere, and how do I prove such a statement?


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