For $x>0$, let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. Then
$$
\lim_{x\to0^+}x\left(\left\lfloor\frac{1}{x}\right\rfloor
+\left\lfloor\frac{2}{x}\right\rfloor+\dots
+\left\lfloor\frac{10}{x}\right\rfloor\right)=\_\_\_
$$
I know , $\lfloor x\rfloor= x $ if $x\in Z$ &
$n$ if $x\notin Z$ and $n \in Z$ and $ n<x<n+1$.
I don't know how to proceed further !!
Best Answer
$\frac{1}{x} -1< \lfloor \frac{1}{x}\rfloor<\frac{1}{x}$
multiplying by x we get $\implies$x[$\frac{1}{x} -1]< x\lfloor \frac{1}{x}\rfloor<x[\frac{1}{x}]$
this gives $1-x< \lfloor \frac{1}{x}\rfloor<1$ taking limit x $\to 0^{+}$ by squeez theorem ,we get $ lim_{x\to 0^{+}} x\lfloor \frac{1}{x}\rfloor=1$
similarly,
$\frac{2}{x} -1< \lfloor \frac{2}{x}\rfloor<\frac{2}{x}$
multiplying by x we get $\implies$x[$\frac{2}{x} -1]< x\lfloor \frac{2}{x}\rfloor<x[\frac{2}{x}]$
this gives $2-x< \lfloor \frac{2}{x}\rfloor<2$
taking limit x $\to 0^{+}$ by squeez theorem ,we get $ lim_{x\to 0^{+}} x\lfloor \frac{2}{x}\rfloor=2$
so this give value of complete expression =55.