Calculating $\lim_{\lambda \rightarrow \infty} T_{\frac{\sin(\lambda x)}{x}}$ in $\mathcal{D}'(\mathbb{R})$ I am looking to calculate 
\begin{equation}
\lim_{\lambda \rightarrow \infty} T_{\frac{\sin(\lambda x)}{x}}
\end{equation}
in the distribution sense. $\frac{\sin(\lambda x)}{x}$ is not locally integrable near $0$, so I thought I'd write: $\forall \varphi \in \mathcal{D(\mathbb{R})}$,
\begin{equation}
\langle T_{\frac{\sin(\lambda x)}{x}}, \varphi \rangle_{\mathcal{D}',\  \mathcal{D}} = \lim_{\varepsilon \rightarrow 0} \int_{\mathbb{R} \setminus [-\varepsilon, \varepsilon] } {\frac{\sin(\lambda x)}{x}} \varphi(x) dx. 
\end{equation} 
I don't know if my proceedings are correct and even if they are, I don't know how to get past this point. I'd appreciate any help. Thanks.
 A: Let $T_{\lambda}$ denote $T_{\frac{\sin(\lambda x)}{x} }$. We have for $\varphi \in D(\Bbb R)$:
$$\langle T_{\lambda}, \varphi\rangle = \int_{-\infty}^{\infty} \frac{\sin (\lambda x)}{x} \varphi(x)dx$$
Write:
$$\varphi(x) = \varphi(0) + x \psi(x), \text{ where } \psi(x) = \int_0^1 \varphi'(tx)dt$$
Then for all $A > 0$ we have:
$$\int_{-A}^A \frac{\sin (\lambda x)}{x} \varphi(x)dx = \varphi(0) \int_{-A}^A \frac{\sin(\lambda x)}{x} dx + \int_{-A}^A \psi(x) \sin(\lambda x) dx\\ = \varphi(0) \int_{-\lambda A}^{\lambda A} \frac{\sin(x)}{x} dx + \int_{-A}^A \psi(x) \sin(\lambda x) dx$$
The first term converges to $\pi\varphi(0)$ as $A \to \infty$. Note that the second term is an absolutely convergent integral (hint: outside the support of $\varphi$, $\psi(x)$ is just $-\varphi(0) \frac1x$ - integrate by parts and take $A$ large enough so that $supp(\varphi) \subset (-A,A)$). We can then write:
$$\langle T_{\lambda}, \varphi \rangle = \pi \varphi(0) + \int_{-\infty}^{\infty} \psi(x) \sin(\lambda x)dx)$$
By the Riemann-Lebesgue lemma:
$$\lim_{\lambda \to \infty}\int_{-\infty}^{\infty} \psi(x) \sin(\lambda x)dx = 0$$
Therefore $\lim_{\lambda \to \infty}\langle T,\varphi\rangle = \pi \varphi(0) = \langle \pi \delta_0, \varphi\rangle$
Showing that $\pi \delta_0$ is the distributional limit of $T_{\lambda}$.
