=x→0−lim∣x∣x([x]+∣x∣)sin[x]
=x→0−lim−xx(−1−x)sin(−1) [x]=−1,∀x∈[−1,0)
=x→0−lim−(1+x)sin1 sin(−θ)=−sinθ
=−sin1.
JEE Main 2019 — Mathematics Calculus
For each x∈R, let [x] be the greatest integer less than or equal to x. Then x→0−lim∣x∣x([x]+∣x∣)sin[x] is equal to
Held on 9 Jan 2019 · Verified 6 Jul 2026.
1
0
−sin1
sin1
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