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

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

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Let see what happens when $x\to \infty$. We have $\ln(1+y)= y-\frac{y^2}{2}+\frac{y^3}{3}+O(y^4)$ as $y\to 0$ hence $$ f(x)= \left(e^x-e^{2x}+e^{4x}\left(e^{-x}-\frac{e^{-2x}}{2}+\frac{e^{-3x}}{3}+O(e^{-4x})\right)^2\right)^{1/3} $$ that is $$ f(x)= \left(e^x-e^{2x}+e^{4x}\left(e^{-2x}-e^{-3x}+\frac{11}{12}e^{-4x}+O(e^{-5x})\right)\right)^{1/3} $$ therefore $$ f(x)= \left(\frac{11}{12}+O(e^{-x})\right)^{1/3}. $$

In particular, $$\lim_{x\to \infty}f(x)=\left(\frac{11}{12}\right)^{1/3}.$$

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  • $\begingroup$ What do you mean by $ O(y^{4}) $? $\endgroup$ – ztefelina Dec 23 '16 at 23:27
  • $\begingroup$ @ztefelina Given functions $f,g: \mathbf{R}\to \mathbf{R}$, we write $f(x)=O(g(x))$ as $x\to \infty$ as a shorthand for $|f(x)| \le C |g(x)|$, for some $C>0$ and all sufficiently large $x$. $\endgroup$ – Paolo Leonetti Dec 23 '16 at 23:30
  • $\begingroup$ Thank you! Unfortunately, I cannot understand how you obtained that $ \left ( e^{-x}-\frac{e^{-2x}}{2}+\frac{e^{-3x}}{3}+O(e^{-4x}) \right )^{2}=e^{-2x}-e^{-3x}+\frac{11}{12}e^{-4x}+O(e^{-5x}) $. $\endgroup$ – ztefelina Dec 23 '16 at 23:39
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    $\begingroup$ @ztefelina What do you do if you have to compute manually $(a+b+c+d)^2$? Try it from the biggest to lowest terms :) $\endgroup$ – Paolo Leonetti Dec 23 '16 at 23:40
  • $\begingroup$ But is it true that $ \lim_{x\rightarrow -\infty }e^{4x}ln^{2}(1+e^{-x})=0 $, as I stated before? $\endgroup$ – ztefelina Dec 23 '16 at 23:54

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