Let
$\hat{f_\epsilon}: \xi \mapsto \exp(-\epsilon |\xi|) \frac{\sin(|\xi|t)}{|\xi| t}$
denote to the Fourier transform of $f$. How do I see
$\hat{f_\epsilon}$ converges uniformly on $\mathbb{R}^n$ to $\hat{f}=\frac{\sin(|\xi|t)}{|\xi|t}$ as $\epsilon \to 0$ ?
$\hat{f_\epsilon} \to \hat{f}$ in the tempered distributions $S'(\mathbb{R}^n)$
${f_\epsilon} \to {f}$ in $S'(\mathbb{R}^n)$?
Who can help me?