Question

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

Answer 1

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]}