Let f(x)=x ($[\frac{1}{x}]+[\frac{2}{x}]+....+[\frac{8}{x}]$) for x$\neq$ 0 and f(x)=$9k$ for x=0 then value of K for which the function is continous

Question

Let f(x) = x ($ [\frac{1}{x}] + [\frac{2}{x}] + [\frac{3}{x}]....+[\frac{8}{x}] $ ) for x$\neq$ 0 and f(x) = $9k$ for x = 0

then the value of k for which the function is continous at x=0 is ([.] denotes greatest integer function)

MY ATTEMPT : we have to find $\lim_{x\to 0}$ f(x) . I have taken x= $ \frac{1}{t}$ so when x$\to$ 0, t$\to$$\infty$

so the expression changes to $$ f(x) = \frac{([t] + [2t] + [3t]....+[8t])}{t} $$

then $\lim_{t\to \infty}$ f(x) looks like $ \frac{\infty}{\infty} $ form but how to use the L-hospital rule with having greatest integer function I am struck here

Please Help Me

Answer 1

I must say $[t]\sim t$ when $t\to\infty$.