limit x->infinity x/e^x
Let's evaluate the limit of xex when x+∞
The limit of ex is +∞ when x+∞
For xex, we encounter the indeterminate form +∞+∞ as x tends to +∞
Let's apply L'Hôpital's rule (limit of a/b = limit of a'/b') by differentiating numerator and denominator:
Let's perform the differentiation: (x)'
1
Let's perform the differentiation: (ex)'
ex
Using those derivatives, we obtain a new fraction whose limit can be computed:
The limit of 1ex is 0⁺ when x+∞
So the limit of xex is 0⁺ when x+∞
Answer:   0