Calculate with Taylor's series the following limit:
$$\lim \limits_{x \to \infty}x-x^2\ln\left(1+\frac{1}{x}\right)$$
As I know, I should open an expansion around the point $a=0$, which means using Maclaurin series, but I saw that what's inside the function $ln$ is not defined in $0$. what should I do? Im I thinking in the right direction? I would like to have a detailed clarification about the technique required to solve this question.
Note: without using L'Hospital's rule.
We have
$$\lim_{x \to \infty}x-x^2\ln\left(1+\frac{1}{x}\right)=\lim_{x \to \infty}x-x^2\left(\frac{1}{x}-\frac1{2x^2}+o\left(\frac1{x^2}\right)\right)=\frac12$$
In order to use Taylor series take $y = \frac{1}{x}$, then $y \to 0$ as $x \to \infty$. Now at a neighborhood of $0$ we have the following
$$\require{cancel}\lim_{y \to 0} \frac{1}{y} - \frac{1}{y^2} \ln (1 + y) = \lim_{y \to 0} \frac{1}{y} - \frac{1}{y^2} \Bigg(y - \frac{y^2}{2} + O(y^3)\Bigg) =\lim_{y \to 0} \color{#05f}{\cancel{\frac{1}{y}}} - \color{#05f}{\cancel{\frac{1}{y}}} + \frac{1}{2} + O(y) = \color{#f05}{\frac{1}{2}}$$
An asymptotic expansion for $\displaystyle f(x)=x-x^2\ln(1+\frac{1}{x})$ is $$f(x)=\sum_{n=2}^{\infty}\frac{(-1)^n}{nx^{n-2}}.$$
