# Find the asymptotes

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 $\lim_{x\to\infty }f(x)$ ?

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