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

Correct answer: D. is equal to 120 We know that (1)/(x) - 1 Hence, (1 + 2 + 3 . . . + 15)/(x) - 15 ⇒ _ x → 0 ^ + x([(1)/(x)] + [(2)/(x)] + . ... + [(15)/(x)]) = 15((16)/(2)) = 120

JEE Main 20188 April

Exam: JEE Main
Year: 2018
Date: 8 April
Subject: Mathematics
Chapter: Limits

JEE Main 2018 · 8 April · Mathematics · Limits

For each t ∈ R, let [t] be the greatest integer less than or equal...

For each

t \in R,

let \left[t\right]

be the greatest integer less than or equal to

. Then

\lim_{x \to {0}^{+}}x\left(\left[\dfrac{1}{x}\right] + \left[\dfrac{2}{x}\right] + \ldots + \left[\dfrac{15}{x}\right]\right)

Options

Adoes not exist $\left(inR\right)$
Bis equal to $0$
Cis equal to $15$
Dis equal to $120$

Correct answer

D. is equal to $120$

Step-by-step solution

We know that

$\dfrac{1}{x} - 1 < \left[\dfrac{1}{x}\right] \le \dfrac{1}{x}$

Hence,

$\dfrac{1 + 2 + 3 . . . + 15}{x} - 15 < \left[\dfrac{1}{x}\right] + \left[\dfrac{2}{x}\right] + .... + \left[\dfrac{15}{x}\right] \le \dfrac{1 + 2 + ... + 15}{x}$

$⇒ \lim_{x \to {0}^{+}}x\left(\left[\dfrac{1}{x}\right] + \left[\dfrac{2}{x}\right] + . ... + \left[\dfrac{15}{x}\right]\right) = 15\left(\dfrac{16}{2}\right)$

$= 120$

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