1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $\displaystyle \lim_{h\to 0}\frac{1}{h}-\frac{1}{h^2}\ln(1+h)$ $\endgroup$ – Galc127 Feb 16 '15 at 12:00
2
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

We have

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

$\endgroup$
  • $\begingroup$ Well, the part that I didn't know how to do is finding the expansion itself. can u elaborate? $\endgroup$ – Firas Ali Abdel Ghani Feb 16 '15 at 12:03
  • 1
    $\begingroup$ 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 $f$ $$f(t)=f(0)+tf'(0)+\frac{t^2}{2!}f''(0)+\cdots$$ $\endgroup$ – user63181 Feb 16 '15 at 12:08
  • $\begingroup$ Cool. Thanks Sami. $\endgroup$ – Firas Ali Abdel Ghani Feb 16 '15 at 12:12
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Feb 16 '15 at 12:16
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.