For each $t \in \mathbf{R}$, let $[t]$ be the greatest integer less than or equal to $t$. Then $\lim _{x \rightarrow 0+} x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots . .+\left[\frac{15}{x}\right]\right)$

(A) does not exist (in $\mathbf{R}$).

JEE Main · 2018 · 08 Apr 2018 · Offline · Limits

For each

t \in R

, let

[t]

be the greatest integer less than or equal to

t

. Then

lim_{x \to 0 +} x ([\frac{1}{x}] + [\frac{2}{x}] + \dots .. + [\frac{15}{x}])

Answer: (D)

Write each greatest-integer term as

[\frac{k}{x}] = \frac{k}{x} - {\frac{k}{x}}

, so that

x \sum_{k = 1}^{15} [\frac{k}{x}] = \sum_{k = 1}^{15} k - x \sum_{k = 1}^{15} {\frac{k}{x}}

. Each fractional part stays between

0

and

1

, so the second sum is bounded and

x

times it vanishes as

x \to 0^{+}

. What survives is

\sum_{k = 1}^{15} k = \frac{15 \times 16}{2} = 120

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f^{'} (3) = 0$ and $f^{''} (3) = 4$ .
Then …
JEE Main 2026View solution →
If $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$ , then $a^{2} + b^{2} + c^{2}$ is equal to :
JEE Main 2026View solution →
The value of
$lim_{x \to 0} \frac{l o g _{e} ( s e c ( e x ) \cdot s e c ( e ^{2} x ) \cdot \dots \cdot s e c ( e ^{10} x ) )}{e ^{2} - e ^{2 c o s x}}$
is equal to
JEE Main 2026View solution →
If $lim_{x \to 2} \frac{s i n ( x ^{3} - 5 x ^{2} + a x + b )}{( x - 1 - 1 ) l o g _{e} ( x - 1 )} = m$ , then $a + b + m$ is equal to:
JEE Main 2026View solution →
The product of all possible values of $α$ , for which $lim_{x \to 0} (\frac{1 - c o s ( α x ) c o s (( α + 1 ) x ) c o s (( α + 2 ) x )}{s i n ^{2} (( α + 1 ) x )}) = 2$ , is :
JEE Main 2026View solution →
Let $f (x) = lim_{y \to 0} \frac{( 1 - c o s ( x y )) t a n ( x y )}{y ^{3}}$ . Then the number of solutions of the equation $f (x) = sin x$ , $x \in R$ is:
JEE Main 2026View solution →