Theorem: N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence

thumb600pxThe graph of the $\ds \frac n \pi \frac 1 {1 + n^2 x^2}$ delta sequence. As $n$ grows, the graph becomes thinner and taller. The area under each graph is equal to $1$.
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\map {\delta_n} x := \frac n \pi \frac 1 {1 + n^2 x^2}$
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where:
:$\phi \in \map \DD \R$ is a test function
:$\delta$ is the Dirac delta distribution
:$\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.

Proof:
Let $\map g x = \map \phi x - \map \phi 0$.
Then:
:$\ds \int_{- \infty}^\infty \map \phi x \map {\delta_n} x \rd x = \map \phi 0 + \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x$
Let $A \in \R_{> 0}$.
Then:
{{begin-eqn}}
{{eqn | l = \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x
      | r = \int_{- \infty}^{- A} \map g x \map {\delta_n} x \rd x + \int_A^\infty \map g x \map {\delta_n} x \rd x + \int_{- A}^A \map g x \map {\delta_n} x \rd x
      | c = Sum of Integrals on Adjacent Intervals for Integrable Functions
}}
{{eqn | o = =:
      | r = I_1 + I_2 + I_3
}}
{{end-eqn}}
Let:
:$\ds \max_{x \mathop \in \closedint {-A} A} \size {\map g x} := \map M A$
Then:
{{begin-eqn}}
{{eqn | l = I_3
      | o = \le
      | r = \int_{-A}^A \size {\map g x} \map {\delta_n} x \rd x
}}
{{eqn | o = \le
      | r = \map M A \int_{-A}^A \frac n {\pi \paren {1 + n^2 x^2} } \rd x
}}
{{eqn | r = \map M A \frac 2 \pi \map \arctan {n A}
      | c = Primitive of Reciprocal of x squared plus a squared
}}
{{eqn | o = \le
      | r = \map M A
      | c = $\ds \forall x \in \R_{\ge 0} : 0 \le \map \arctan x \le \frac \pi 2$
}}
{{end-eqn}}
We have that $\map g 0 = 0$.
By definition, $\phi$ is a smooth real function on $\R$.
By Differentiable Function is Continuous, $\map g x$ is continuous at $x = 0$. 
Furthermore:
{{begin-eqn}}
{{eqn | l = \lim_{A \mathop \to 0} \map M A 
      | r =  \lim_{A \mathop \to 0} \max_{x \mathop \in \closedint {-A} A} \size {\map g x}
}}
{{eqn | r = \size {\map g 0}
}}
{{eqn | r = 0
}}
{{end-eqn}}
By definition of the limit of a real function:
:$\forall \epsilon' \in \R_{>0} : \exists \delta \in \R_{>0}: \forall A \in \R_{> 0}: 0 < A < \delta \implies \map M A < \epsilon'$
Let $\ds \epsilon' = \frac \epsilon 2$.
It follows that:
:$\ds \forall \epsilon \in \R_{> 0} : \exists A \in \R_{> 0} : I_3 \le \map M A < \frac \epsilon 2$
Suppose $A$ is such that the above inequality holds.
By definition, $\map \phi x$ is bounded.
Then:
{{begin-eqn}}
{{eqn | l = \size {\map g x}
      | r = \size {\map \phi x - \map \phi 0}
}}
{{eqn | o = < 
      | r = \size {\map \phi x} + \size {\map \phi 0}
      | c = Triangle Inequality for Real Numbers
}}
{{end-eqn}}
It follows that:
:$\exists b \in \R_{> 0} : \forall x \in \R : \size {\map g x} < b$ 
Then:
{{begin-eqn}}
{{eqn | l = \size {I_1 + I_2} 
      | r = \size { \int_{- \infty}^{-A} \map g x \map {\delta_n} x \rd x + \int_A^\infty \map g x \map {\delta_n} x \rd x}
}}
{{eqn | o = \le
      | r = \size {\int_{- \infty}^{-A} \size {\map g x} \map {\delta_n} x \rd x + \int_A^\infty \size {\map g x} \map {\delta_n} x \rd x}
}}
{{eqn | o = \le
      | r = \size { b \int_{- \infty}^{-A} \map {\delta_n} x \rd x + b \int_A^\infty \map {\delta_n} x \rd x}
}}
{{eqn | o = \le 
      | r = b \size {\int_{- \infty}^{-A} \map {\delta_n} x \rd x + \int_A^\infty \map {\delta_n} x \rd x} 
}}
{{eqn | r = \frac b \pi \size {\bigintlimits {\map \arctan {n x} } {- \infty} {-A} + \bigintlimits {\map \arctan {n x} } A \infty }
      | c = Primitive of Reciprocal of x squared plus a squared
}}
{{eqn | r = \frac b \pi \size {- \map \arctan {n A} + \frac \pi 2 + \frac \pi 2 - \map \arctan {n A} }
}}
{{eqn | r = b \size {1 - \frac 2 \pi \map \arctan {n A} }
}}
{{end-eqn}}
With the number $A$ fixed: 
:$\ds \lim_{n \mathop \to \infty} \frac 2 \pi \map \arctan {n A} = 1$.
By Squeeze Theorem and for a fixed $A$ we have:
:$\ds \lim_{n \mathop \to \infty} \size {I_1 + I_2} = 0$
By definition of the limit of a real sequence: 
:$\ds \forall \overline \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \size {I_1 + I_2} \le b \size {1 - \frac 2 \pi \map \arctan {n A} } < \overline \epsilon$.
Let $\ds \overline \epsilon = \frac \epsilon 2$.
Then:
:$\ds \forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \size {I_1 + I_2} < \frac \epsilon 2$.
Let $A$ and $N$ be such that the above inequalities for $I_3$ and $I_1 + I_2$ hold.
Then:
{{begin-eqn}}
{{eqn | l = \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} 
      | o = \le 
      | r = \size {I_1 + I_2 + I_3} 
}}
{{eqn | o = \le 
      | r = \size {I_1 + I_2} + \size {I_3} 
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | o = < 
      | r = \epsilon
}}
{{end-eqn}}
To sum up:
:$\ds \forall \epsilon \in \R_{>0} : \exists N \in \R_{>0} : \forall n \in \N_{>0} : \forall n > N \implies \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} < \epsilon$
By definition of the limit of a real sequence:
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x = 0$
However:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x
      | r = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0 \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \rd x
      | c = Sum Rule for Limits of Real Functions
}}
{{eqn | r = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x - \map \phi 0
}}
{{eqn | r = 0
}}
{{eqn | ll = \leadsto
      | l = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map \phi x \map {\delta_n} x \rd x
      | r = \map \phi 0
}}
{{end-eqn}}
{{qed}}