This is a question related to L'hopital's rule.
lim x -> -infinity of ((x)(e^x))
This thing is weird because I apply l'hopital's rule yet I never receive the correct answer which is supposedly = 0. It just stays in the form [-infinity * -infinity].
Anyone can help out?
Thought I answered this
-x/e^x
e^x >> x for large x
How did you get that Damon? I thought I could only reorder question as e^x/(1/x) and apply hopital rule from there
e^x as x --> big negative is 1/e^big negative
e^-a = 1/e^a
I mean
e^x as x --> big negative is 1/e^big positive
To apply L'Hopital's rule, we need to evaluate the limit of the ratio of the derivatives of the numerator and denominator. Let's go step by step:
1. Start by differentiating the numerator and the denominator:
Numerator: (d/dx)(x) = 1
Denominator: (d/dx)(e^x) = e^x
2. Now, we can rewrite the expression and reapply L'Hopital's rule:
lim x -> -infinity of (x * e^x) = lim x -> -infinity of (1 / e^x)
3. Differentiate the numerator and the denominator again:
Numerator: (d/dx)(1) = 0
Denominator: (d/dx)(e^x) = e^x
4. Rewrite the expression and reapply L'Hopital's rule:
lim x -> -infinity of (1 / e^x) = lim x -> -infinity of (0 / e^x)
At this point, we have a limit of the form "0 / e^x" as x approaches negative infinity. However, it is important to note that when we have a zero in the numerator and a non-zero value in the denominator (such as e^x in this case), the limit is equal to 0.
Therefore, we can conclude that the limit lim x -> -infinity of ((x)(e^x)) is indeed 0.