solution xe^1 as x goes to 0 and (x-1/x)^x as x goes to infinity using L'hopitals rule?

Answer 1

First of all, I think this is a math problem.
Second of all, none of these ever needs using L'hopitals rule.
xe^1 =ex
ex goes to 0 when x goes to 0.
As x goes to infinity, (x -1/x)^x goes to x^x which goes to infinity.