How do you find all the asymptotes for #R(x)=(3x+5) / (x-6)#?

1 Answer
Oct 18, 2015

Horizontal: #y=3#, for #x\to \pm \infty#.
Vertical: #\pm \infty#, for #x \to 6^\pm#.

Explanation:

The function is defined where the denominator is not zero, which means that all points are ok except for #x=6#. So, the domain is #(-infty,6) cup (6,infty)#.

The possible asymptotes are thus found computing the limits of #R(x)# as #x \to -\infty#, #x\to 6^{-}#, #x \to 6^{+}#, #x \to infty#.

  • #lim_{x \to -\infty} R(x) = lim_{x \to -\infty} (x(3+5/x))/(x(1-6/x))#
    You can simplify the #x#'s, and both #5/x# and #-6/x# tend to zero. So, the limit is #3#.

  • For the limits in #6#, from left and right, we have that the denominator tends to zero, so the function will tend to positive or negative infinity, depending on the signs: if #x\to 6^{-}#, then the denominator tends to zero from negative values, and the numerator is positive, so the limit is #-infty#. On the contrary, if #x\to 6^{+}#, both numerator and denominator are positive, so the limit is #infty#.

  • The limit for #x\to\infty# requires exactly the same calculations as the one for #x\to-\infty#, so the asymptote will be #y=3# again.

  • Since the function has horizontal asymptotes, there can't be oblique ones.