α=x→∞lim((1−ee)(e1−1+xx))x(1∞ form )∴α=eL Where L=x→∞limx((1−ee)(e1−1+xx)−1)⇒L=x→∞lim(1−ee)x(e1−1+xx−(e1−e))⇒L=1−eex→∞limx(1−1+xx)⇒L=1−eex→∞limx+1x⇒ L=1−ee⋅1⇒ L=1−ee∴α=e1−ee⇒logα=1−ee∴ Required value =1+1−ee1−ee=e
If x→∞lim((1−ee)(e1−1+xx))x=α, then the value of 1+logeαlogeα equals :
Held on 22 Jan 2025 · Verified 6 Jul 2026.
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