Math asymptotes?

Question

How do you solve:

1) f(x)=ln(1+e^x)

2) f(x)=(x-1)^2(x+1)^2/x^4

3) f(x)=2/sqrt(x^2-4)

I would really appreciate the help. Thnx

Trying to find the horizontal and vertical asymptoes.

Answer 1

Rule of thumb: To solve for the vertical asymptotes, you need to be dealing with functions with fractions, and set the denominator to 0.

To solve for the horizontal asymptotes, you need to find the limit of the function as x approaches infinity, and the limit of the function as x approaches negative infinity.

Let's use these to guide us.

1) f(x) = ln(1 + e^x).

Since we're not dealing with any fractions, there aren't any vertical asymptotes. If we take the limit as x approaches infinity, it doesn't converge to a number, so there aren't any horizontal asymptotes.

2) f(x) = [(x-1)^2 * (x+1)^2]/x^4

To solve for the vertical asymptotes, make the bottom equal to 0.
x^4 = 0 implies x = 0.
Therefore, x = 0 is a vertical asymptote.

To solve for the horizontal asymptotes, you must first expand the numerator.

f(x) = [(x^2 - 2x + 1) (x^2 + 2x + 1)] / x^4
f(x) = (x^4 + 2x^3 + x^2 - 2x^3 - 4x^2 - 2x + x^2 + 2x + 1)/x^4

Group like terms,

f(x) = (x^4 - 2x^2 + 1)/x^4

At this point, we take the limit as x approaches infinity.

lim (x -> infinity, (x^4 - 2x^2 + 1)/x^4 )

And we divide everything by x^4

lim (x -> infinity, (1 - 2/x^2 + 1/x^4)/1 ) =
1 + 0 + 0 = 1

Therefore, our horizontal asymptote is y = 1. If we test the limit as x approaches -infinity, we'll get the same result.

3.
f(x) = 2/sqrt(x^2 - 4)

Vertical asymptote: make the bottom equal to zero.

sqrt(x^2 - 4) = 0
x^2 - 4 = 0
x = -2, 2

Therefore, our vertical asymptotes are x = -2 and x = 2.

To solve for the horizontal asymptote, we find the limit as x approaches infinity.

f(x) = 2/sqrt(x^2 - 4)

Change 2 into sqrt(4)

f(x) = sqrt(4)/sqrt(x^2 - 4)

And now, put under one square root.

f(x) = sqrt ( 4/(x^2 - 4))

Now, we solve it like we did the other equation, resulting in
sqrt (lim (x -> infinity, [4/x^2]/[1 - 4/x^2]) =
sqrt (0/1) = 0

Therefore, y = 0 is a horizontal asymptote.