how do you determine vertical/horizontal asymptotes of xe^-x?

Answer 1

f(x) = xe^(-x)

Remember it is always proper mathematical etiquette to have no negative exponents. So we bring it down as a denominator.

f(x) = x/(e^x)

To solve for the vertical asymptotes, we need to ask ourselves what makes f(x) undefined. The simple answer to that is, what makes the denominator equal to 0, i.e.

e^x = 0

This has no solution, so there are no vertical asymptotes.

To solve for the horizontal asymptotes, you take to take the limit as x approaches infinity.

lim (x => infinity, x/(e^x))

To solve this, we need to use L'Hospital's rule, since we get the form [infinity/infinty]. Thus, we take the derivative of the numerator and the derivative of the denominator.

lim (x => infinity, 1/(e^x)), to which we get the form [1/infinity], which is equal to 0.

We similarly do the same thing for -infinity, but that limit does not exist.

Therefore, there aren't any vertical asymptotes, but the horizontal asymptotes is y = 0.

Answer 2

DEFINITION OF VERTICAL ASYMPTOTE
The line x = m is a vertical asymptote if lim x-->m f(x) = ∞ or -∞.

DEFINITION OF HORIZONTAL ASYMPTOTE
The line y = n is a horizontal asymptote if lim x-->∞ or -∞ f(x) = n.

Therefore, to find vertical asymptotes, we must find all the lines in which the function y = xe^-x approaches but never really reaches. It means that we must find values in which the function is undefined. Since there is no restriction, then we can say that there are no horizontal asymptotes.

To find horizontal asymptotes, find the limit of the function as x approaches ±∞. Therefore,
lim x-->±∞ xe^-x = lim x-->±∞ x/e^x

By L'Hopital's rule,
= lim x-->±∞ 1/e^x

By simplifying,
= lim x-->±∞ (1/e)^x

Since 1/e < 1, then we can say that the limit tends to
= 0

Therefore, the line y = 0 is a horizontal asymptote of the equation, and there are no vertical asymptotes.

^_^

Answer 3

the vertical asymptotes are lines of the form x=a, where
a is a number where f(x) is NOT defined and
lim x->a f(x) is not defined either.
in this case f(x) =xe^-x is always defined, so there are NO vertical asymptotes.

for the horizontal asymptotes you need to check:
lim x->infinity xe^-x = 0 so there is a horizontal asymptote: y=0
and
lim x-> - infinity xe^-x = - infinity, so there is only one horizontal asymptote....