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

##### 1 Answer

The first step is always to simplify the complex fraction to see if we may obtain the function in the form

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

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

There will be vertical asymptotes whenever the denominator equals

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

There is therefore only one vertical asymptote at

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

**Calculus**

We divide each term by the highest degree of the entire function, which will be

#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

#y = (0 + 0)/1#

#y = 0#

This confirms what we found above without any calculus.

**Practice exercises**

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

Hopefully this helps, and good luck!