Skip to main content

For each t∈ R, let [t] be the greatest integer less than or equal to t. Thenundersetx→ 0^+ limx([ 1 x]+[ 2 x]+… +[ 15 x])

JEE Main 2018 — Mathematics Calculus

2018mcqhard

For each t∈R,t\in R,t∈R, let [t][t][t] be the greatest integer less than or equal to ttt. Thenlimx→0+x([1x]+[2x]+…+[15x])\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}x([\frac{1}{x}]+[\frac{2}{x}]+\ldots +[\frac{15}{x}])x→0+lim​x([x1​]+[x2​]+…+[x15​])

Official previous-year question

Held on 8 Apr 2018 · Verified 6 Jul 2026.

Options

  1. A

    does not exist (inR)(\mathrm{in}R)(inR)

  2. B

    is equal to 000

  3. C

    is equal to 151515

  4. D

    is equal to 120120120

Did you get this right?

Sign in to track your attempts and accuracy.

Your note

Sign in to keep a private note on this question. Nothing you write is ever public.

JEE Main Calculus in other years

More JEE Main Calculus questions

Other Mathematics topics for JEE Main

Keep practicing Calculus

Work through every JEE Main Calculus PYQ, year by year.

Browse all Mathematics topics