Dirac delta sequences

No. For instance, take the function $\delta_n(x) = 1$ when $x \in [n, n+1]$ and $0$ otherwise.

EDIT: The following conditions will give the conclusion you want: $\delta_n(x) \geq 0$, $\int_\mathbb{R} \delta_n(x) = 1$, $\int_{-\epsilon}^{\epsilon} \delta_n(x) dx \to 1$ for all $\epsilon > 0$. I'm pretty sure other combinations of conditions will work, but I'm not sure what necessary and sufficient conditions are. That's actually an interesting question.

Here's the argument: Let $$ E_n = \int_\mathbb{R} \phi(x) \delta_n(x) \, dx - \phi(0) = \int_\mathbb{R} (\phi(x)-\phi(0)) \delta_n(x) \, dx $$ using the second condition. Then $$ |E_n| \leq \int_{\epsilon}^{\epsilon} |\phi(x)-\phi(0)|\delta_n(x) \, dx + \int_{\mathbb{R} \setminus [-\epsilon, \epsilon]} |\phi(x)-\phi(0)| \delta_n(x) \, dx $$ or $$ |E_n| \leq \sup_{x \in [-\epsilon, \epsilon]} |\phi(x)-\phi(0)| \int_{-\epsilon}^\epsilon \delta_n(x) \, dx + 2 \sup_{x \in \mathbb{R}} |\phi(x)| \int_{\mathbb{R} \setminus [-\epsilon, \epsilon]} \delta_n(x) \, dx. $$ By the last (and second) condition, $$ \limsup_{n \to \infty} |E_n| \leq \sup_{x \in [-\epsilon, \epsilon]} |\phi(x)-\phi(0)|. $$ Take $\epsilon \to 0$ and you're done.

You could set have $ \delta_n(x) = n/2 $ if $x\in (0,1/n]$ or $x\in (1,1+1/n]$. Then $$ \lim \int_{-\infty}^\infty \phi(x)\delta_n(x)dx = \frac12(\phi(0)+\phi(1)) $$ for all continuous functions $\phi$.