 # Find the asymptotes of f:R rightarrow R, f(x)= root(3)(ex−e2x+e4xln2(1+e−x)). I found that y=0 is an asymptote when x rightarrow −infinity, but how do I calculate limx rightarrow infinity f(x) ? Braylon Lester 2022-07-23 Answered
Find the asymptotes
Find the asymptotes of $f:\mathbb{R}\to \mathbb{R},f\left(x\right)=\sqrt{{e}^{x}-{e}^{2x}+{e}^{4x}{\mathrm{ln}}^{2}\left(1+{e}^{-x}\right)}.$
I found that y=0 is an asymptote when $x\to -\mathrm{\infty }$, but how do I calculate $\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)$ ?
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Let see what happens when $x\to \mathrm{\infty }$. We have $\mathrm{ln}\left(1+y\right)=y-\frac{{y}^{2}}{2}+\frac{{y}^{3}}{3}+O\left({y}^{4}\right)$ as $y\to 0$ hence
$f\left(x\right)={\left({e}^{x}-{e}^{2x}+{e}^{4x}{\left({e}^{-x}-\frac{{e}^{-2x}}{2}+\frac{{e}^{-3x}}{3}+O\left({e}^{-4x}\right)\right)}^{2}\right)}^{1/3}$
that is
$f\left(x\right)={\left({e}^{x}-{e}^{2x}+{e}^{4x}\left({e}^{-2x}-{e}^{-3x}+\frac{11}{12}{e}^{-4x}+O\left({e}^{-5x}\right)\right)\right)}^{1/3}$
therefore
$f\left(x\right)={\left(\frac{11}{12}+O\left({e}^{-x}\right)\right)}^{1/3}.$
In particular,
$\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)={\left(\frac{11}{12}\right)}^{1/3}.$