Braylon Lester

2022-07-23

Find the asymptotes

Find the asymptotes of $f:\mathbb{R}\to \mathbb{R},f(x)=\sqrt[3]{{e}^{x}-{e}^{2x}+{e}^{4x}{\mathrm{ln}}^{2}(1+{e}^{-x})}.$

I found that y=0 is an asymptote when $x\to -\mathrm{\infty}$, but how do I calculate $\underset{x\to \mathrm{\infty}}{lim}f(x)$ ?

Find the asymptotes of $f:\mathbb{R}\to \mathbb{R},f(x)=\sqrt[3]{{e}^{x}-{e}^{2x}+{e}^{4x}{\mathrm{ln}}^{2}(1+{e}^{-x})}.$

I found that y=0 is an asymptote when $x\to -\mathrm{\infty}$, but how do I calculate $\underset{x\to \mathrm{\infty}}{lim}f(x)$ ?

Brendon Bentley

Beginner2022-07-24Added 11 answers

Let see what happens when $x\to \mathrm{\infty}$. We have $\mathrm{ln}(1+y)=y-\frac{{y}^{2}}{2}+\frac{{y}^{3}}{3}+O({y}^{4})$ as $y\to 0$ hence

$f(x)={({e}^{x}-{e}^{2x}+{e}^{4x}{({e}^{-x}-\frac{{e}^{-2x}}{2}+\frac{{e}^{-3x}}{3}+O({e}^{-4x}))}^{2})}^{1/3}$

that is

$f(x)={({e}^{x}-{e}^{2x}+{e}^{4x}({e}^{-2x}-{e}^{-3x}+\frac{11}{12}{e}^{-4x}+O({e}^{-5x})))}^{1/3}$

therefore

$f(x)={(\frac{11}{12}+O({e}^{-x}))}^{1/3}.$

In particular,

$\underset{x\to \mathrm{\infty}}{lim}f(x)={\left(\frac{11}{12}\right)}^{1/3}.$

$f(x)={({e}^{x}-{e}^{2x}+{e}^{4x}{({e}^{-x}-\frac{{e}^{-2x}}{2}+\frac{{e}^{-3x}}{3}+O({e}^{-4x}))}^{2})}^{1/3}$

that is

$f(x)={({e}^{x}-{e}^{2x}+{e}^{4x}({e}^{-2x}-{e}^{-3x}+\frac{11}{12}{e}^{-4x}+O({e}^{-5x})))}^{1/3}$

therefore

$f(x)={(\frac{11}{12}+O({e}^{-x}))}^{1/3}.$

In particular,

$\underset{x\to \mathrm{\infty}}{lim}f(x)={\left(\frac{11}{12}\right)}^{1/3}.$

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