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Question Number 125292 by Kurbanklichevs last updated on 09/Dec/20
limx→0(x+1)n+1−(n+1)(x+1)+nx2n∈NCalculate.(WithoutaL′Hopital′srule)
Answered by Dwaipayan Shikari last updated on 09/Dec/20
limx→0(1+x)n+1−(n+1)(x+1)+nx2limx→01+(n+1)x+n(n+1)2!x2−(n+1)(x+1)+nx2limx→01−n−1−n+n(n+1)2!x2x2=n(n+1)2
Commented by Kurbanklichevs last updated on 10/Dec/20
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