# I was trying to find asymptotes for a function f ( x ) = x e <mrow class="MJ

I was trying to find asymptotes for a function $f\left(x\right)=x{e}^{\frac{1}{x-2}}$
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luisjoseblash2
When x is large, let
$\frac{1}{x-2}=t\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=2+\frac{1}{t}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}{e}^{\frac{1}{x-2}}=\frac{\left(2t+1\right)}{t}{e}^{t}$
Now, using Taylor around t=0
$\frac{\left(2t+1\right)}{t}{e}^{t}=\frac{1}{t}+3+\frac{5t}{2}+\frac{7{t}^{2}}{6}+O\left({t}^{3}\right)$
Back to x
$x\phantom{\rule{thinmathspace}{0ex}}{e}^{\frac{1}{x-2}}=\frac{6{x}^{3}-18{x}^{2}+15x+1}{6\left(x-2{\right)}^{2}}$
Long division
$x\phantom{\rule{thinmathspace}{0ex}}{e}^{\frac{1}{x-2}}=x+1+\frac{5}{2x}+O\left(\frac{1}{{x}^{2}}\right)$
which gives not only the slant asymptote but also shows how the function does approach it.
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Layla Velazquez
We have:
$\underset{x\to +\mathrm{\infty }}{lim}f\left(x\right)=\underset{x\to +\mathrm{\infty }}{lim}x\cdot {e}^{\frac{1}{x-2}}=+\mathrm{\infty }$
We notice that ${e}^{\frac{1}{x-2}}\to 1$, so it's true the following asymptotic relation:
$f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sim \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x$
when $x\to +\mathrm{\infty }$. This is a necessary but not sufficient condition for the existence of the asympot.
We are searching for line of the type $y=x+k,\phantom{\rule{thinmathspace}{0ex}}k\in \mathbb{R}$
$\underset{x\to +\mathrm{\infty }}{lim}f\left(x\right)-x=\underset{x\to +\mathrm{\infty }}{lim}x\cdot \left({e}^{\frac{1}{x-2}}-1\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sim \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{x\to +\mathrm{\infty }}{lim}x\cdot \frac{1}{x-2}=1$
Here, we have used a very important asymptotic relation:
${e}^{f\left(x\right)}-1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sim \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\to {x}_{0}$
With a function $f\left(x\right)$ such that $f\left(x\right)\to 0$ when $x\to {x}_{0}\in \overline{\mathbb{R}}$
Here, I put $f\left(x\right)=\frac{1}{x-2}$ that is such that $f\left(x\right)\to 0$ when $x\to +\mathrm{\infty }$
In conclusion, the asympot is:
$y=x+1$