What are the asymptotes and removable discontinuities, if any, of #f(x)=4-1/(x+5) +1/x#?

1 Answer
Feb 19, 2018

No removanble discontinues, vertical asymptotes at #x=0# and #x=-5# and horizontal asymptotes at #y=4#

Explanation:

As #f(x)=4-1/(x+5)+1/x=(4x(x+5)-x+x+5)/(x(x+5))#

= #(4x^2+20x+5)/(x(x+5)#

As #x# or #x+5# is not a factor of #4x^2+20x+5#, there are no removanble discontinues.

Vertical asymptotes are at #x=0# and #x+5=0# i.e. #x=-5#, because as #x->0# or #x->-5#, #f(x)->+-oo#, depending on whether we approach from left or right.

Now we can write #f(x)=(4x^2+20x+5)/(x(x+5)#

= #(4x^2+20x+5)/(x^2+5x)#

= #(4+20/x+5/x^2)/(1+5/x)#

Hence as #x->oo#, #f(x)->4#

and we have horizontal asymptote #y=4#

graph{4-1/(x+5)+1/x [-21.92, 18.08, -5.08, 14.92]}