$\lim _{n \to \infty} \int_{0}^1 f \left(\frac{x}{n}\right) dx =0$ find $f(0) ?$ let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $$\lim _{n \to \infty} \int_{0}^1 f \left(\frac{x}{n}\right) dx =0$$
Then what we can say about $f(0) \ $ ??
how  to approach this problem . Any hint .
 A: You have to couple the continuity of the function $f$ with the convergence of the series:
Let $\varepsilon>0$. Thus, there exists a $\delta>0$ such that $-\varepsilon<f(y)-f(0)<\varepsilon$ for all $-\delta<y<\delta$. 
Since in the integral, $0<x<1$ and therefore $0<x/n<1/n$ there exists an $N$ such that $x/n<\delta$ for all $n\geq N$. This implies that:
$f(0)-\varepsilon<\int_{0}^1 f \left(\frac{x}{n}\right) dx<f(0)+\varepsilon$ for all $n\geq N$.
Which finally yields $\lim_{n\to\infty} \int_{0}^1 f \left(\frac{x}{n}\right) dx = f(0)$.
A: We must have $f(0)=0$.
Suppose wlog that $f(0)>0$. Fix an interval $[a,b]$ with $a>0$ and $f(0)\in [a,b]$.
Since $f$ is continuous, there is $\delta>0$ such that $f(x) \in [a,b]$ whenever $|x| < \delta$.
But then for all $n$ such that $1/n < \delta$ we have
$$
\int_{0}^1 f \left(\frac{x}{n}\right) \, dx \ge \int_{0}^1 a \, dx = a
$$ 
Therefore,
$$
\lim _{n \to \infty} \int_{0}^1 f \left(\frac{x}{n}\right) \, dx \ge a > 0
$$
A: The key here is the Fundamental Theorem of Calculus.
Consider the substitution $x=nt$ and then the given limit condition is equivalent to $$\lim_{n\to\infty} n\int_{0}^{1/n}f(t)\,dt=0$$ On the other hand since $f$ is continuous at $0$ we have via Fundamental Theorem of Calculus $$\lim_{h\to 0}\frac {1}{h}\int_{0}^{h}f(t)\,dt=f(0)$$ Putting $h=1/n,n\in\mathbb {N} $ we get $$\lim_{n\to\infty} n\int_{0}^{1/n}f(t)\,dt=f(0)$$ It should now be clear that $f(0)=0$. 
