real analysis - There is some $NinBbb Z^+$ such that $|f(x_n)| geq frac{|f(x)|}{2}$ for all $ngeq N$

Let $f:\Bbb R \to \Bbb R$ be a continuous function. Let $\{x_n\}_{n=1}^\infty$ be a convergent sequence in $\Bbb R$ with $\lim \limits_{n\to\infty} x_n = x$ and $f(x)\ne 0$

I want to prove that there is some $N\in\Bbb Z^+$ such that $|f(x_n)| \geq \frac{|f(x)|}{2}$ for all $n\geq N$

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I think I should do this by proving that for a continuous $f$, $x_n\to x \implies f(x_n)\to f(x)$, and then using that to prove the statement.

Is this the correct approach?