If $f(x)=\left\{\begin{array}{r}x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+...+\left[\frac{n}{x}\right]\right), &x ≠ 0 \\ k \quad \quad ~~~~~~~~~~, & x=0\end{array}\right.$ and $n \in N$. Then the value of k for which f(x) is continuous at x = 0, is

$\frac{n(n+1)}{2}$

For any x > 0, we know that

$\frac{1}{x}-1<\left[\frac{1}{x}\right] \leq \frac{1}{x} \Rightarrow 1-x<x\left[\frac{1}{x}\right] \leq 1$

$\frac{2}{x}-1<\left[\frac{2}{x}\right] \leq \frac{2}{x} \Rightarrow 2-x<x\left[\frac{2}{x}\right] \leq 2$

$\vdots \quad \vdots \quad \vdots \quad \vdots \quad \vdots$

$\frac{n}{x}-1<\left[\frac{n}{x}\right] \leq \frac{n}{x} \Rightarrow n-x<x\left[\frac{n}{x}\right] \leq n$

∴  $\frac{n(n+1)}{2}-n x<x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+...+\left[\frac{n}{x}\right]\right) \leq \frac{n(n+1)}{2}$

But,

$\lim\limits_{x \rightarrow 0}\left\{\frac{n(n+1)}{2}-n x\right\}=\frac{n(n+1)}{2}$  and,  $\lim\limits_{x \rightarrow 0} \frac{n(n+1)}{2}=\frac{n(n+1)}{2}$

Therefore, by Sandwhich theorem, we obtain

$\lim\limits_{x \rightarrow 0} x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+...+\left[\frac{n}{x}\right]\right)=\frac{n(n+1)}{2}$