What are the asymptote(s) and hole(s), if any, of # f(x) = x/(x-1)-(x-1)/x#?

1 Answer
Jun 12, 2017

Answer:

#x=0# is an asymptote.

#x=1# is an asymptote.

Explanation:

First, let's simplify this so that we have a single fraction that we can take the limit of.

#f(x) = (x(x))/((x-1)(x)) - ((x-1)(x-1))/(x(x-1))#

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

#f(x) = (2x-1)/((x-1)(x))#

Now, we need to check for discontinuities. This is just anything that will make the denominator of this fraction #0#. In this case, to make the denominator #0#, #x# could be #0# or #1#. So let's take the limit of #f(x)# at those two values.

#lim_(x->0)(2x-1)/(x(x-1)) = (-1)/(-1*0) = +-oo#

#lim_(x->1)(2x-1)/(x(x-1)) = 3/(1*0) = +-oo#

Since both of these limits tend towards infinity, both #x=0# and #x=1# are asymptotes of the function. There are therefore no holes in the function.