$\lim_{n \rightarrow \infty } \int_0^1 \frac{nf(x)}{1+ n^{2} x^{2} }dx = \frac{ \pi }{2} f(0)$ for continuous $f$. If f is a continuous function on $[0,1]$, then show that $$\lim_{n \rightarrow  \infty }  \int_0^1  \frac{nf(x)}{1+ n^{2} x^{2}  }dx   = \frac{ \pi }{2} f(0)$$
Can anybody help me to solve this? I tried but i have no idea about how to prove this.
Thanks in advance!
 A: Put $I = \int_{0}^1 \frac{nf(x)}{1+n^2x^2} dx = I_1 + I_2$ where
$I_1 = \int_{0}^{n^{-\frac13} } \frac{nf(x)}{1+n^2x^2} dx$ and $I_2 = \int_{n^{-\frac13} }^1 \frac{nf(x)}{1+n^2x^2} dx$.
Further $|f(x)| \le M$ because $f$ is continious. We have
$$|I_2| \le  \int_{n^{-\frac13} }^1 \bigg|\frac{nf(x)}{1+n^2x^2} \bigg| dx \le  \int_{n^{-\frac13} }^1 \frac{nM}{1+n^2x^2} dx \le  \int_{n^{-\frac13} }^1 \frac{nM}{1+n^2(n^{-\frac13})^2} dx  $$
$$=\frac{nM}{1+n^2(n^{-\frac13})^2} \cdot  (1  - n^{-\frac13}) = o(1). $$
Put $I_3 = \int_{0}^{n^{-\frac13} } \frac{nf(0)}{1+n^2x^2} dx$. We have
$$|I_1 - I_3| = \bigg| \int_{0}^{n^{-\frac13} } \frac{n(f(x)-f(0))}{1+n^2x^2} dx \bigg| \le  \int_{0}^{n^{-\frac13} } \frac{n |f(x)-f(0)|}{1+n^2x^2} dx $$
$$ \le \sup_{ x \in [0, n^{-\frac13}]} |f(x)-f(0)| \cdot \int_{0}^{n^{-\frac13} } \frac{n}{1+n^2x^2} dx  = o(1) \cdot \int_{0}^{n^{-\frac13} } \frac{d(nx)}{1+(nx)^2}  $$
$$ =   o(1) \cdot  \arctan (n \cdot n^{-\frac13}) = o(1) \cdot O(1) = o(1). $$
Hence $$I = I_1 + I_2 = I_3 + (I_1 - I_3) + I_2 = I_3 + o(1) + o(1)$$ where
$$I_3 = f(0) \int_{0}^{n^{-\frac13} } \frac{d(nx)}{1+n^2x^2} =  f(0)  \cdot  \arctan (n \cdot n^{-\frac13}) = f(0) \frac{\pi}2 +o(1). $$
So $I_1 \to  f(0) \frac{\pi}2$ as $n \to \infty$.
Addition:
As Oolong milk tea said above, we may solve the problem much easier if it's allowed to use dominated convergence theorem: $I = \int_{[0,n]} \frac{f(\frac{y}n)}{1+y^2}dy = o(1) + \int_{[0,\infty]} \frac{f(\frac{y}n)}{1+y^2}dy = o(1) + \int_{[0,\infty]} \frac{f(0)}{1+y^2}dy$ where $f(x) = f(1)$ for $x > 1$.
A: Let's see what properties does the kernel have:
$$K_n(x) = \frac{2}{\pi}\cdot \frac{n}{1 + n^2 x^2}$$


*$$K_n(x) \ge 0$$


*$$\lim_{n\to \infty} \int_0^1 K_n(x) dx = 1$$
Indeed, we have $\int_0^1 \frac{n}{1+ n^2 x^2} dx= \int_{0}^n \frac{dx}{1+x^2} \to \frac{\pi}{2}$ as $n\to \infty$.


*For every $0< \delta < 1$, we have

$$\lim_{n\to \infty} \int_{\delta}^1 K_n(x) dx = 0$$
Similar check
Using these properties, let's prove the statement.
We have
$$\int_0^1 K_n(x) f(x) dx - f(0) = \int_0^1 K_n(x)( f(x) - f(0) )dx  + (\int_0^1 K_n(x)dx  -1 )\cdot f(0) $$
The second term approaches $0$ by 1. Let's show that the first also converges to $0$. We have
$$ \int_0^1 K_n(x) (f(x) - f(0)) dx = \int_0^{\delta} \cdots + \int_{\delta}^1 \cdots $$
Take $\epsilon> 0$ arbitrary. Choose $\delta> 0$ such that $|f(x)-f(0)|< \epsilon$ for $x\in (0, \delta)$. Now the first part on RHS above is in absolute values $\le \epsilon \int_0^1 |K_n(x)| dx <I \cdot \epsilon$, while the second part in absolute value $\le 2M \int_{\delta}^1 |K_n(x)| dx $, and can be made small as $n\to 0$. We are done.
Let's note that we can use some weaker conditions for kernels $K_n(x)$
0'. $$\int_0^1 |K_n(x)| dx < I$$
for some $I$
1'. $$\lim_{n\to \infty} \int_0^1 K_n(x) dx = 1 $$
2'  For every $0<\delta < 1$ we have
$$\lim_{n\to \infty} \int_{\delta}^1 |K_n(x)| dx = 0$$
Also, it is enough that $f$ is bounded on $[0,1]$, and continuous at $0$.
