Finding slant asymptote of a function $x*e^{1/(x-2)}$ 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.
 A: 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.
A: We have:
$$\lim_{x\to +\infty}f(x)=\lim_{x\to +\infty}x\cdot e^{\frac{1}{x-2}}=+\infty$$
We notice that $e^{\frac{1}{x-2}}\to 1$, so it's true the following asymptotic relation:
$$f(x)\,\,\sim\,\, x$$
when $x\to +\infty$. This is a necessary but not sufficient condition for the existence of the asympot.
We are searching for line of the type $y=x+k, \, k\in \mathbb{R}$:
$$\lim_{x\to +\infty}f(x)-x=\lim_{x\to +\infty}x\cdot\left(e^{\frac{1}{x-2}}-1\right)\,\,\sim\,\,\lim_{x\to +\infty}x\cdot\frac{1}{x-2}=1$$
Here, we have used a very important asymptotic relation:
$$e^{f(x)}-1\,\,\sim\,\, f(x)\,\,\,x\to x_0$$
With a function $f(x)$ such that $f(x)\to 0$ when $x\to x_0\in\bar{\mathbb{R}}$.
Here, I put $f(x)=\frac{1}{x-2}$ that is such that $f(x)\to 0$ when $x\to +\infty$.
In conclusion, the asympot is:
$$y=x+1$$
A: You have\begin{align}\lim_{x\to\infty}x\exp\left(\frac1{x-2}\right)-x&=\lim_{x\to\infty}x\left(\exp\left(\frac1{x-2}\right)-1\right)\\&=\lim_{x\to\infty}(x-2)\left(\exp\left(\frac1{x-2}\right)-1\right)+\\&\qquad+\lim_{x\to\infty}2\left(\exp\left(\frac1{x-2}\right)-1\right)\\&=\lim_{x\to\infty}(x-2)\left(\exp\left(\frac1{x-2}\right)-1\right)\\&=\lim_{x\to\infty}x\left(\exp\left(\frac1x\right)-1\right)\\&=\lim_{x\to0^+}\frac{e^x-1}x\\&=\exp'(0)\\&=1.\end{align}Therefore, $x+1$ is an asymptote of your function.
