Clarification on the proof of the Riemann-Lebesgue lemma in Folland's Real Analysis: Why does $f\in C^1\cap C_c\implies\hat{f}\in C_0$?

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This question concerns one detailthe given proof of the Riemann-Lebesgue lemma in Folland's Real Analysis - Modern Techniques and Their Applications (2nd edition), pp. 249. The claim is that $\mathcal{F}\left(L^1\left(\mathbb{R}^n\right)\right)\subset C_0\left(\mathbb{R}^n\right)$. The author begins the proof as follows:

By e.), if $f\in C^1\cap C_c$, then $|\xi|\hat{f}(\xi)$ is bounded and hence $\hat{f}\in C_0$.

where the part e.) is

If $f\in C^k$, $\partial^\alpha f\in L^1$ for $|\alpha|\leq k$, and $\partial^\alpha f\in C_0$ for $|\alpha|\leq k - 1$, then $\hat{\left(\partial^\alpha f\right)(\xi)} = \left(2\pi i \xi\right)^\alpha \hat{f}(\xi)$

and $C_c$ is the set of continuous functions with compact support and $C_0$ is the set of continuous functions vanishing at the infinity.

What I can't recall/grasp is that why does the Fourier transform of $f\in C^1\cap C_c$ vanish at infinity?