Find the asymptote of function with Taylor formula.

Question

Give function

$f(x)=\frac{x^{x+1}}{(x+1)^{x}}.$

Find it's asymptote with Taylor formula.

I got $f(x)=x\exp{(-x^2+\frac{3}{2}x+o(x^3))}.$But I can't see any asymptotes in here.

Answer 1

$$f(x)=\frac{x^{x+1}}{(x+1)^{x}}\quad \implies \quad\log(f(x))=(x+1)\log(x)-x \log(x+1)$$

For large $x$, by Taylor, $$\log(f(x))=\log (x)-1+\frac{1}{2 x}-\frac{1}{3 x^2}+O\left(\frac{1}{x^3}\right)$$

Taylor again $$f(x)=e^{\log(f(x))}=\frac{x}{e}+\frac{1}{2 e}-\frac{5}{24 e x}+O\left(\frac{1}{x^2}\right)$$ which even gives a curved asymptote and which at least tell how the function approaches its linear asymptote.

Answer 2

I have a suggestion for you. Rewrite function as below $$f(x)=\frac{x^{x+1}}{(x+1)^{x}}=\\\frac{x^{x}.x}{(x+1)^{x}}\\ =\frac{x}{\frac{(x+1)^{x}}{x^{x}}}\\=\frac{x}{(1+\frac 1x)^{x}}$$ then maybe you think twice $(1+\frac 1x)^{x} \to e$ as $x$ tends to $\infty$ so $f(x) \sim \frac {x}{e}$