limit x->infinity x/e^x
1) Compute the limit of xex as x → +∞
ⓘ A number greater than 1 raised to a power that grows without bound grows without bound, a number between 0 and 1 tends to 0
The limit of ex is +∞ when x → +∞
For xex, we encounter the indeterminate form +∞+∞ as x tends to +∞
2) Let's apply L'Hôpital's rule (limit of a/b = limit of a'/b') by differentiating numerator and denominator:
3) Let's differentiate: ddx(x)
ⓘ The derivative of x is 1: ddx(x) = 1
1
4) Let's perform the differentiation: ddx(ex)
ⓘ The derivative of exp(x) is exp(x): ddx(ex) = ex
ex
With the derivatives of the numerator and the denominator, we obtain a new fraction whose limit can be determined:
ⓘ A number divided by a quantity that grows without bound tends to 0
The limit of 1ex is 0⁺ when x → +∞
So the limit of xex is 0⁺ when x → +∞
▶Limit:   0