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

Winigefx
2022-06-30
Answered

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

You can still ask an expert for help

luisjoseblash2

Answered 2022-07-01
Author has **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

Answered 2022-07-02
Author has **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:

$y=x+1$

asked 2022-06-06

How can I find the constants A and B if

$\underset{x\to \mathrm{\infty}}{lim}(\sqrt{{x}^{2}+x}-Ax-B)=0\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}?$

$\underset{x\to \mathrm{\infty}}{lim}(\sqrt{{x}^{2}+x}-Ax-B)=0\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}?$

asked 2021-10-20

Evaluate $\underset{x\to \mathrm{\infty}}{lim}\frac{5{x}^{2}-3x}{9+{e}^{x}}$

asked 2020-12-02

Evaluate the limit

$\underset{x\to \mathrm{\infty}}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}$

asked 2020-12-24

Find the limits

$\underset{y\to 2}{lim}\frac{y+2}{{y}^{2}+5y+6}$

asked 2021-01-30

Evaluate the following limits: $\underset{y\to 0}{lim}\frac{1-\mathrm{cos}(7y)}{2y}$

asked 2022-04-29

limit $\underset{t\to 0}{lim}\frac{\mathrm{sin}3t}{\mathrm{tan}2t}$

I am struggling with this question. I have attempted it, but I keep getting stuck on the following step:

$\underset{t\to 0}{lim}\frac{1}{2}\frac{\mathrm{sin}3t\mathrm{cos}2t}{\mathrm{sin}t\mathrm{cos}t}$

I am struggling with this question. I have attempted it, but I keep getting stuck on the following step:

asked 2022-07-26

Find $L=\underset{n\to \mathrm{\infty}}{lim}\frac{2{n}^{2}+6n-1}{4{n}^{2}+2n+1}$. then determine and find .