Find the asymptotes

Question

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) $ ?

Answer 1

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