6
$\begingroup$

How to calculate the following limit?

$\lim_{x \to 0}\left(\frac1{x} + \frac{\ln(1-x)}{x^2}\right)$

  • 0
    @Adrian: Well, I think it's his wish whether to upvote an answer or not.2011-05-19

3 Answers 3

3

using the series $\log(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}$ for $-1\leq x<1$ we have $\frac{1}{x}+\frac{\log(1-x)}{x^2}=-\frac{1}{x^2}\sum_{n=2}^{\infty}\frac{x^n}{n}$ so the limit as $x\to0$ is $-\frac{1}{2}$

12

You have \begin{eqnarray*} \lim_{x\to 0} \left(\frac{1}{x}+\frac{\ln(1-x)}{x^2}\right) &=& \lim_{x\to 0} \frac{x+\ln(1-x)}{x^2}, \end{eqnarray*} note that $ \lim_{x\to 0} x+\ln(1-x)=0,\: \lim_{x\to 0} x^2= 0, $ then by the L'Hospital's rule $\begin{align*} \lim_{x\to 0} \left(\frac{1}{x}+\frac{\ln(1-x)}{x^2}\right)&= \lim_{x\to 0} \frac{\frac{d}{dx}(x+\ln(1-x))}{\frac{d}{dx}x^2}\\ &= \lim_{x\to 0} \frac{1-\frac{1}{1-x}}{2x}\\ &= \lim_{x\to 0}\frac{\frac{1-x - 1}{1-x}}{2x}\\ &= \lim_{x\to 0} \frac{\frac{x}{x-1}}{2x}\\ &= \lim_{x\to 0}\frac{1}{2(x-1)}\\ &= -\frac{1}{2}. \end{align*}$

  • 0
    Someone can help me?2011-05-06
10

A not so elegant way is to represent $\log(1-x)$ as a power series for $|x| < 1$ i.e. $\log(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \ldots$ Plug this in to get $\frac1{x} + \frac{\log(1-x)}{x^2} = \frac1{x} - \frac{x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots}{x^2} = -\frac1{2} - \frac{x}{3} - \frac{x^2}{4} - \cdots$ Hence, the desired limit is $-\frac1{2}$

  • 1
    I agree with @user6312 and I think you confound *elegant* with *based on tricks*. Your solution is automatic, unimaginative, simple... hence it is the best one. Had you replaced the irrelevant parts of your three infinite developments by $+o(x^2)$ (twice) and by $+o(1)$ (once), it would have been perfect. :-)2011-05-07