How do you find the vertical, horizontal or slant asymptotes for #y = e^x (1-x^2)#?

Question

James Allen · Accepted Answer

Horizontal asymptote at #y=0#

Firstly, there are no singularities in this function (there is nowhere where we would have to "divide by 0"). As such there are no vertical asymptotic.

Lets look at the case where:

#x->+oo# The function then becomes:

#e^x(1-x^2)-> -e^x x^2#

as the #x^2 # term dominates.

This increases non-linearly and as such will not be asymptotic. So there are no asymptotes going when #x->+oo#.

Now let's look at when #x-> -oo#. The function becomes:

#e^x(1-x^2)->e^x x^2#

again.

However in this case, #x# is negative so the exponential product: #e^x# will get closer and closer to #0#, and will do much faster than the #x^2# product can "pull the function up."

As such there will be a horizontal asymptote at #y=0#:

graph{e^x(1-x^2) [-10, 10, -5, 5]}

Lily Anderson · Answer

undefined To find the vertical asymptote, we need to identify any values of $ x $ that make the denominator equal to zero. However, in this case, there is no denominator, so there are no vertical asymptotes.

To find the horizontal or slant asymptotes, we examine the behavior of the function as $ x $ approaches positive or negative infinity. Since $ e^x $ grows without bound as $ x $ approaches infinity, and $ 1 - x^2 $ approaches negative infinity as $ x $ approaches either positive or negative infinity, there are no horizontal or slant asymptotes. Therefore, this function does not have horizontal, slant, or vertical asymptotes.