limit x->infinity x^2/e^x
1) Compute the limit of x2ex as x → +∞
ⓘ A quantity that grows without bound raised to a positive power still grows without bound, to a negative power it tends to 0
The limit of x2 is +∞ when 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 x2ex, we encounter the indeterminate form +∞+∞ as x tends to +∞
2) We can use L'Hôpital's rule (limit of a/b = limit of a'/b') by computing the derivatives of the numerator and the denominator:
3) Let's perform the differentiation: ddx(x2)
ⓘ The derivative of x^n is n*x^(n-1): ddx(x2) = 2x
2x
4) Let's differentiate: 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 quantity that grows without bound multiplied by a quantity not tending to 0 still grows without bound, the signs giving +∞ or -∞
The limit of 2x is +∞ when x → +∞
For 2xex, we encounter the indeterminate form +∞+∞ as x tends to +∞
5) Let's apply L'Hôpital's rule (limit of a/b = limit of a'/b') by differentiating numerator and denominator:
6) Let's perform the differentiation: ddx(2x)
ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u': ddx(2x) = 2
2
7) 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 2ex is 0⁺ when x → +∞
The limit of 2xex is 0⁺ when x → +∞
So the limit of x2ex is 0⁺ when x → +∞
▶Limit:   0