x→10lim2x−68x3−βxx2(αx)+(γ−1)(1+1x2)=3x→0lim(2−β)x−34x3(γ−1)+(γ−1)x2+αx3=3γ−1,β=2,4−3α=+3⇒α=−4β+γ−α=7
JEE Main 2025 — Mathematics Calculus
For α,β,γ,∈R, if x→0limsin2x−βxx2sinαx+(γ−1)ex2=3, then β+γ−α is equal to:
Held on 2 Apr 2025 · Verified 6 Jul 2026.
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