Winigefx

2022-06-30

I was trying to find asymptotes for a function $f(x)=x{e}^{\frac{1}{x-2}}$

luisjoseblash2

Beginner2022-07-01Added 16 answers

When x is large, let

$\frac{1}{x-2}=t\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=2+\frac{1}{t}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}{e}^{\frac{1}{x-2}}=\frac{(2t+1)}{t}{e}^{t}$

Now, using Taylor around t=0

$\frac{(2t+1)}{t}{e}^{t}=\frac{1}{t}+3+\frac{5t}{2}+\frac{7{t}^{2}}{6}+O\left({t}^{3}\right)$

Back to x

$x\phantom{\rule{thinmathspace}{0ex}}{e}^{\frac{1}{x-2}}=\frac{6{x}^{3}-18{x}^{2}+15x+1}{6(x-2{)}^{2}}$

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.

$\frac{1}{x-2}=t\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=2+\frac{1}{t}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}{e}^{\frac{1}{x-2}}=\frac{(2t+1)}{t}{e}^{t}$

Now, using Taylor around t=0

$\frac{(2t+1)}{t}{e}^{t}=\frac{1}{t}+3+\frac{5t}{2}+\frac{7{t}^{2}}{6}+O\left({t}^{3}\right)$

Back to x

$x\phantom{\rule{thinmathspace}{0ex}}{e}^{\frac{1}{x-2}}=\frac{6{x}^{3}-18{x}^{2}+15x+1}{6(x-2{)}^{2}}$

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.

Layla Velazquez

Beginner2022-07-02Added 11 answers

We have:

$\underset{x\to +\mathrm{\infty}}{lim}f(x)=\underset{x\to +\mathrm{\infty}}{lim}x\cdot {e}^{\frac{1}{x-2}}=+\mathrm{\infty}$

We notice that ${e}^{\frac{1}{x-2}}\to 1$, so it's true the following asymptotic relation:

$f(x)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sim \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x$

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 $y=x+k,\phantom{\rule{thinmathspace}{0ex}}k\in \mathbb{R}$

$\underset{x\to +\mathrm{\infty}}{lim}f(x)-x=\underset{x\to +\mathrm{\infty}}{lim}x\cdot ({e}^{\frac{1}{x-2}}-1)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sim \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{x\to +\mathrm{\infty}}{lim}x\cdot \frac{1}{x-2}=1$

Here, we have used a very important asymptotic relation:

${e}^{f(x)}-1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sim \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}f(x)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\to {x}_{0}$

With a function $f(x)$ such that $f(x)\to 0$ when $x\to {x}_{0}\in \overline{\mathbb{R}}$

Here, I put $f(x)=\frac{1}{x-2}$ that is such that $f(x)\to 0$ when $x\to +\mathrm{\infty}$

In conclusion, the asympot is:

$y=x+1$

$\underset{x\to +\mathrm{\infty}}{lim}f(x)=\underset{x\to +\mathrm{\infty}}{lim}x\cdot {e}^{\frac{1}{x-2}}=+\mathrm{\infty}$

We notice that ${e}^{\frac{1}{x-2}}\to 1$, so it's true the following asymptotic relation:

$f(x)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sim \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x$

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 $y=x+k,\phantom{\rule{thinmathspace}{0ex}}k\in \mathbb{R}$

$\underset{x\to +\mathrm{\infty}}{lim}f(x)-x=\underset{x\to +\mathrm{\infty}}{lim}x\cdot ({e}^{\frac{1}{x-2}}-1)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sim \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{x\to +\mathrm{\infty}}{lim}x\cdot \frac{1}{x-2}=1$

Here, we have used a very important asymptotic relation:

${e}^{f(x)}-1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sim \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}f(x)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\to {x}_{0}$

With a function $f(x)$ such that $f(x)\to 0$ when $x\to {x}_{0}\in \overline{\mathbb{R}}$

Here, I put $f(x)=\frac{1}{x-2}$ that is such that $f(x)\to 0$ when $x\to +\mathrm{\infty}$

In conclusion, the asympot is:

$y=x+1$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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