Given L=x→0+limx−x3cos−1(x−[x]2)⋅sin−1(x−[x]2)
∵x∈[0,1)⇒[x]=0
Then L=x→0+lim(1−x2)cos−1x×x→0+limxsin−1x
L=2π×x→0+lim1−x21(Apply L’Hospital rule for 00 form)
L=2π
JEE Main 2021 — Mathematics Calculus
The value of x→0+limx−x3cos−1(x−[x]2)⋅sin−1(x−[x]2), where [x] denotes the greatest integer ≤x is:
Held on 17 Mar 2021 · Verified 6 Jul 2026.
π
0
4π
2π
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