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

The first step is always to simplify the complex fraction to see if we may obtain the function in the form #f(x) = (g(x))/(h(x))#.

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

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

There will be vertical asymptotes whenever the denominator equals #0#.

#x^2 = 0 -> x = 0#

There is therefore only one vertical asymptote at #x= 0#.

Now we calculate the horizontal asymptotes. I will write an answer that doesn't involve calculus and then an answer that does involve calculus.

No Calculus

By the rules of asymptotes, there will be a horizontal asymptote at #y = 0# (because the highest degree of the denominator is higher than the numerator).

Calculus

We divide each term by the highest degree of the entire function, which will be #x^2# and take the limit as #x# approaches positive infinity.

#y = lim_(x-> oo) (x/x^2 + 1/x^2)/(x^2/x^2)#

#y = lim_(x-> oo) (1/x + 1/x^2)/1#

By the identity #lim_(x-> oo) 1/x = 0#, we have

#y = (0 + 0)/1#

#y = 0#

This confirms what we found above without any calculus.

Practice exercises

#1#. Describe asymptotes, both horizontal and vertical, in the following function.

#g(x) = (x + 3 + 2/(x - 4))/(x - 1)#

#2#. Solutions:

#1#. V.A: #x = 4 and x = 1#. H.A: #y = 1#

Hopefully this helps, and good luck!