Here is an intuition behind the problem:
Suppose that a family of functions $K_n$ has the following property:
- $K_n \geq 0$ and $\int K_n = 1$. That is, $K_n$ has unit mass,
- $\int_{|x|\geq \delta} K_n \to 0$ as $n \to \infty$. That is, the mass of $K_n$ concentrates toward $0$ as $n$ grows.
We can give an intuitive interpretation of these conditions in terms of $K_n(x) \, dx$ as follows: Think of an integral $\int f(x) \, dx $ as the sum of infinitesimal masses $f(x) \, dx$, each of which is located on the interval $[x, x+dx]$. Then
- the total sum the infinitesimal masses $ K_n(x) \, dx $ is equal to one,
- $ K_n(x) \, dx \to 0$ if $x$ is away from $0$ as $n\to\infty$.
Thus we expect that $ K_n(0) \, dx \to 1$ and hence
\begin{align*}
\lim_{n\to\infty} \int f(x) K_n(x) \, dx
&= \lim_{n\to\infty} \sum f(x) K_n(x) \, dx \\
&= \sum f(x) \lim_{n\to\infty} K_n(x) \, dx
= f(0).
\end{align*}
This observation(?) suggests that we should divide the behavior of $K_n$ into
- near-the-origin part where the mass of $K_n$ accumulates and
- away-from-the-origin part where the mass of $K_n$ vanishes.
Now let us return the the question of devising a rigorous proof. Let $f$ be continuous on $[0, 1]$. In particular, $f$ is bounded by some constant $M > 0$ and continuous at $x = 0$. Thus for any $\epsilon > 0$, there exists $\delta > 0$ such that
$$ |x| < \delta \Longrightarrow |f(x) - f(0)| < \epsilon. $$
Now let
$$ K_n(x) = \frac{1}{\tan^{-1}n} \frac{n}{1+n^2 x^2}. $$
Then it is clear that
$$ \int_{0}^{1} K_n(x) \, dx = \frac{1}{\tan^{-1} n} \int_{0}^{n} \frac{dx'}{1+x'^2} = 1$$
and similarly
\begin{align*}
\int_{\delta}^{1} K_n(x) \, dx
&= \frac{1}{\tan^{-1} n} \int_{n\delta}^{n} \frac{dx'}{1+x'^2} \\
&\leq \frac{1}{\tan^{-1} n} \int_{n\delta}^{\infty} \frac{dx'}{1+x'^2}
= \frac{\tan^{-1} 1/(n\delta)}{\tan^{-1} n} \to 0 \quad \text{as } n\to\infty.
\end{align*}
Keeping these observations in mind, we make the following decomposition.
\begin{align*}
\left| \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx - \frac{\pi}{2} f(0) \right|
&\leq \left| \int_{0}^{1} f(x) K_n(x) \, dx - f(0) \right| \tan^{-1}n + \left|\tan^{-1} n - \frac{\pi}{2} \right| \left| f(0) \right|.
\end{align*}
Then dividing the integral term into two parts with one near from the origin and the other away from the origin, we observe that
\begin{align*}
\left| \int_{0}^{1} f(x) K_n(x) \, dx - f(0) \right|
&= \left| \int_{0}^{1} (f(x) - f(0)) K_n(x) \, dx \right| \\
&\leq \int_{0}^{1} \left| f(x) - f(0) \right| K_n(x) \, dx \\
&\leq \int_{0}^{\delta} \left| f(x) - f(0) \right| K_n(x) \, dx + \int_{\delta}^{1} \left| f(x) - f(0) \right| K_n(x) \, dx \\
&\leq \int_{0}^{\delta} \epsilon K_n(x) \, dx + \int_{\delta}^{1} 2M K_n(x) \, dx \\
&\leq \epsilon + 2M \int_{\delta}^{1} K_n(x) \, dx,
\end{align*}
and hence
\begin{align*}
\left| \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx - \frac{\pi}{2} f(0) \right|
&\leq \frac{\pi}{2} \epsilon + \pi M \int_{\delta}^{1} K_n(x) \, dx + \left| f(0) \right| \tan^{-1} \frac{1}{n}.
\end{align*}
Taking $\limsup_{n\to\infty}$, we have
\begin{align*}
\limsup_{n\to\infty} \left| \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx - \frac{\pi}{2} f(0) \right|
&\leq \frac{\pi}{2} \epsilon.
\end{align*}
But since this is true for any $\epsilon > 0$, we must have
\begin{align*}
\limsup_{n\to\infty} \left| \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx - \frac{\pi}{2} f(0) \right| = 0 \quad \Longleftrightarrow \quad \lim_{n\to\infty} \int_{0}^{1} \frac{n f(x)}{1+n^2 x^2} \, dx = \frac{\pi}{2} f(0).
\end{align*}