# Calculate with Taylor's series the following limit: $\lim\limits_{x \to \infty}x-x^2\ln\left(1+\frac{1}{x}\right)$

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.

• Can you expand $\ln\left(1+\frac{1}{x}\right)$ at $x=\infty$? Other way is defining $x=\frac{1}{h}$ and then find the limit $\displaystyle \lim_{h\to 0}\frac{1}{h}-\frac{1}{h^2}\ln(1+h)$ – Galc127 Feb 16 '15 at 12:00

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

$$\require{cancel}\lim_{y \to 0} \frac{1}{y} - \frac{1}{y^2} \ln (1 + y) = \lim_{y \to 0} \frac{1}{y} - \frac{1}{y^2} \Bigg(y - \frac{y^2}{2} + O(y^3)\Bigg) =\lim_{y \to 0} \color{#05f}{\cancel{\frac{1}{y}}} - \color{#05f}{\cancel{\frac{1}{y}}} + \frac{1}{2} + O(y) = \color{#f05}{\frac{1}{2}}$$

We have

$$\lim_{x \to \infty}x-x^2\ln\left(1+\frac{1}{x}\right)=\lim_{x \to \infty}x-x^2\left(\frac{1}{x}-\frac1{2x^2}+o\left(\frac1{x^2}\right)\right)=\frac12$$

• Well, the part that I didn't know how to do is finding the expansion itself. can u elaborate? – Firas Ali Abdel Ghani Feb 16 '15 at 12:03
• You should know the expansion on $0$ of $$\ln(1+t)=t-\frac{t^2}2+\frac{t^3}3+\cdots$$ and to find it apply the Taylor formula on a function $f$ $$f(t)=f(0)+tf'(0)+\frac{t^2}{2!}f''(0)+\cdots$$ – user63181 Feb 16 '15 at 12:08
• Cool. Thanks Sami. – Firas Ali Abdel Ghani Feb 16 '15 at 12:12
• You're welcome. – user63181 Feb 16 '15 at 12:16

An asymptotic expansion for $\displaystyle f(x)=x-x^2\ln(1+\frac{1}{x})$ is $$f(x)=\sum_{n=2}^{\infty}\frac{(-1)^n}{nx^{n-2}}.$$