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) ?

Question

Find the asymptotes
Find the asymptotes of

f : R \to R, f (x) = \sqrt[3]{e^{x} - e^{2 x} + e^{4 x} \ln^{2} (1 + e^{- x})} .

I found that y=0 is an asymptote when

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

that is

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

therefore

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

In particular,

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

Answers (1)