We need to calculate lim(x→∞)−(x+1)(e(x+1)1−1)
=lim(x→∞)(−1)(x+1)1e(x+1)1−1
It is in 00 form after putting ∞ in place of x .So L hospital's rule is applicable here.
=(−1)lim(x→∞)dxd(x+1)1dxd(e(x+1)1−1)
=(−1)lim(x→∞)(x+1)2−1e(x+1)1.(x+1)2−1 (∵dxdvu =v2(v.u′−u.v′)) So (dxdx+11=(x+1)2−1)
( dxdex=ex)
=(−1)lim(x→∞)e(x+1)1
=(−1)e∞1
=(−1)e0
=(−1).1=−1 (Ans)
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