# maths-calculus

find the limit as x approaches infinity, -(x+1) (e^(1/(x+1))-1)

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1. If you let u = -(x+1), you have

u(e^(-1/u) - 1)
= (e^(-1/u) - 1) / (1/u)

Now using l'Hospital's Rule, that is the same as

(e^(-1/u) - 1)(1/u^2) / (1/u^2)
= e^(-1/u) - 1
-> 0-1 = -1

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posted by Steve

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