Find any horizontal asymptote of the function (Please show the work to get the asymptote)

Question

a) f(x) = cos(pi(x)/3x+2)

b)h(x) = arctan(1+e^x)

Answer 1

To find the horizontal asymptotes, we take the limit of the functions as x approaches +∞ and -∞.

(a) The way you have typed the function is as follows:

f(x) = cos(π(x)/3x + 2)

If this is indeed correct, the x's in the numerator and denominator cancel out and the function can be simplified as follows:

f(x) = cos(π/3 + 2) = -0.996

This is just a constant and the graph would be a horizontal line. However, I suspect that maybe a parenthesis was omitted and the correct function is:

f(x) = cos(π(x)/(3x + 2))

If this is the case, now we can examine this function in detail and take the limit as x approaches + and - ∞.

Looking inside the cosine, we have a polynomial of degree 1 (π*x) divided by another polynomial of degree 1 (3x + 2). When the limit of a polynomial divided by another polynomial of the same degree goes to positive and negative infinity, we simply look at the coefficients in front of the leading terms. This is π/3.

cos(π/3) = 1/2, this is the horizontal asymptote at both positive and negative infinity.

We can easily verify this is correct by graphing the function and examining the behavior as x gets very large and very small.

(b) Our function in part (b) is:

h(x) = arctan(1 + e^x)

First lets look at the limit as x approaches +∞. Again, start by examining the part of the function inside the parenthesis. (1 + e^x). e^x is the natural exponential function, and we can see somewhat intuitively that as x gets very large and approaches infinity, that the resulting exponent will also get very large and will approach infinity. Adding +1 to infinity effectively does not make a difference in this problem and we are left with:

arctan(1 + ∞) = arctan(∞)

So we are looking for some unknown angle (in radians) such that tan(x) = ∞. We know that tan = sin/cos, and this ratio will approach ∞ as the denominator approaches zero. This occurs at an angle of π/2.

arctan(∞) = π/2, this is one horizontal asymptote as x approaches positive infinity.

Now let's look at when x approaches -∞.

e^-∞ = 1/(e^∞) = 0

arctan(1 + 0) = arctan(1), we are looking for when sin = cos since we want sin/cos to equal 1

arctan(1) = π/4, this is the other horizontal asymptote as x approaches negative infinity.

Again, we can easily verify these results are correct by graphing the function.