How do you find vertical, horizontal and oblique asymptotes for #f(x)=x(e^(1/x))#?

1 Answer
Nov 5, 2016

Answer:

The oblique asymptote is #y=x+1#
The vertical asymptote is #x=0#

Explanation:

We know the Taylor's series for
#e^x=1+x+x^2/(2!)+x^3/(3!)+......#
#e^(1/x)=1+1/x+1/(x^2*2!)+1/(x^3*3!+.....#
#:.f(x)=xe^(1/x)=x(1+1/x+1/(x^2*2!)+1/(x^3*3!+.....))#
#=x+1+1/(x*2!)+1/(x^2*3!)+...#
therefore #y=x+1# is an oblique asymptote
And #x=0# is a vertical asymptote

graph{(y-xe^(1/x))(y-x-1)=0 [-6.815, 7.235, -2.36, 4.66]}