I was trying to find asymptotes for a function $f(x)=xe^{\frac{1}{x-2}}$.
I calculated vertical asymptote at $x\to 2$.
Slant function is $y=kx+n$.
But when I tried to find slant asymptote, I calculated $\lim_{x\to \infty} f(x)=\frac{xe^{\frac{1}{x-2}}}{x}$, so the result was $k=1$.
Now when I wanted to find $n$, calculating $\lim_{x\to\infty}f(x)=xe^{\frac{1}{x-2}}-x$ because $k=1$, I didn't know how to get to the right answer.
I tried using L'Hopital's rule, but my function kept expanding, so I couldn't get the right answer. Most of the online calculators gave the answer $n = 1$, but none gave me steps to this solution.
Best Answer
When $x$ is large, let $$\frac 1 {x-2}=t \implies x=2+\frac{1}{t}\implies x\,e^{\frac{1}{x-2}}=\frac{ (2 t+1)}{t}e^t$$ Now, using Taylor around $t=0$ $$\frac{ (2 t+1)}{t}e^t=\frac{1}{t}+3+\frac{5 t}{2}+\frac{7 t^2}{6}+O\left(t^3\right)$$ Back to $x$ $$x\,e^{\frac{1}{x-2}}=\frac{6 x^3-18 x^2+15 x+1}{6 (x-2)^2}$$ Long division $$x\,e^{\frac{1}{x-2}}=x+1+\frac{5}{2 x}+O\left(\frac{1}{x^2}\right)$$ which gives not only the slant asymptote but also shows how the function does approach it.