I'm trying to solve this limit without the use of L'Hospital, but I'm doing something wrong. The limit should be:
limx→0(e−x−1x)=−1
My attempted proof:
limx→0(e−x−1x)=limn→∞(e−(1n)−11n)=limn→∞(n⋅(e−(1n)−1)1)=limn→∞(n⋅(e0−1)1)=limn→∞(01)=0
I assume the mistake is that I've used the continuity of the exp function.
Answer
A variation of Bernard's answer:
e−x−1x=1−exxex=−1ex⋅ex−1x→x→0−11⋅(ex)′x=0=−e0=−1
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