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

Nov 5, 2016

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]}