How do you find the asymptotes for #(2x-4)/(x^2-4)#?

2 Answers
Sep 2, 2015

Answer:

Vertical asymptotes: #x=\pm 2#
Horizontal asymptotes: #y=0#.
No oblique asymptotes.

Explanation:

You have vertical asymptotes where the function is not defined, and this function is not defined where its denominator equals zero. So, we have

#x^2-4 =0 \iff x^2=4 \iff x=\pm \sqrt{4}#

which means #x=\pm 2#

As for horizontal asymptotes, you have them if the limits as #x\to\pm\infty# exist and are finite. In your case, both limits exist and equal zero, because the degree of the denominator is greater than the degree of the numerator.

The presence of horizontal asymptotes excludes the one of oblique asymptotes.

Sep 2, 2015

Answer:

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

has vertical asymptote at #x=-2#, horizontal asymptote #y = 0# and removable singularity at #x=2#

Explanation:

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

with exclusion #x != 2#

When #x = -2#, the denominator is zero and the numerator is non-zero. So there is a vertical asymptote (simple pole) at #x = -2#.

When #x = 2#, both the numerator and denominator of #f(x)# are zero, so #f(2)# is undefined, but this is a removable singularity. Both the left and right limits exist at #x=2# and are equal to #1/2#.

As #x->+-oo#, #2/(x+2) -> 0#, so #f(x) -> 0#.

So #f(x)# has a horizontal asymptote #y = 0#.

graph{(2x-4)/(x^2-4) [-10, 10, -5, 5]}

It is possible for a function to have two horizontal asymptotes, or two oblique asymptotes, or one of each.

For example, the function #arctan(x)# has horizontal asymptotes #y = pi/2# and #y = -pi/2#.

graph{arctan(x) [-10, 10, -5, 5]}

Consider also

#g(x) = (x+abs(x))/2+1/(x^2+1)#

This slightly messy function has horizontal asymptote #y=0# and oblique asymptote #y = x#

graph{(x+abs(x))/2+1/(x^2+1) [-8.89, 8.885, -4.434, 4.45]}