Finding limit in D'(R)

Question

How to find $\lim_{\varepsilon\rightarrow 0+} f_\varepsilon$ in $D'(R)$, if $f_\varepsilon(x)=\frac{\sin{\varepsilon x}}{x}$?

Thanks in advance.

Answer 1

By substitution $\varepsilon x\mapsto y$ the expression becomes for a test function $\phi$ $$ \int_{\varepsilon\operatorname{supp }\phi} \frac{\sin y}{y} \phi(\frac{y}{\varepsilon}) dy $$ for all $\varepsilon<1$. Noting the bounded integrand is bounded, this expression converges to $0$ as $\varepsilon\operatorname{supp }\phi$ shrinks to $\{0\}$. To summarize: $$ \lim_{\varepsilon\to 0_+} \frac{\sin \varepsilon x}{x} = 0 \text{ in } \mathcal{D}'(\mathbb{R}). $$