Using L'Hôpital's rule, evaluate lim of xe^(-x) as x approaches infinity
X=Lim xe^(-x)
x->∞
=Lim x/e^x
x->∞
L'hôpital's rule is applicable if direct evaluation results in an indeterminate form, which is the case here.
To apply the rule, differentiate both numerator and denominator to give
X=Lim 1/e^x
x->∞
=0
To evaluate the limit of xe^(-x) as x approaches infinity using L'Hôpital's rule, we can differentiate both the numerator and denominator and continue doing so until we can evaluate the limit.
Let's start by differentiating xe^(-x). The derivative of xe^(-x) with respect to x is (1-x)e^(-x), using the product rule.
Now, we have (1-x)e^(-x) as the numerator and e^(-x) as the denominator. Let's differentiate both of these functions.
The derivative of (1-x)e^(-x) is (-1)e^(-x)-e^(-x), which simplifies to -e^(-x)-e^(-x)=-2e^(-x).
The derivative of e^(-x) is -e^(-x).
Now, we have -2e^(-x) as the numerator and -e^(-x) as the denominator. Let's differentiate again.
The derivative of -2e^(-x) is 2e^(-x).
The derivative of -e^(-x) is e^(-x).
Now, we have 2e^(-x) as the numerator and e^(-x) as the denominator. Since we have reached a constant numerator and denominator, further differentiation is not required.
Taking the limit as x approaches infinity, we get:
lim(x approaches infinity) (xe^(-x)) = lim(x approaches infinity) (2e^(-x)/e^(-x)) = lim(x approaches infinity) 2 = 2.
Therefore, the limit of xe^(-x) as x approaches infinity is 2.