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How to calculate the following limit?

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

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    Jichao, I see that you have asked 28 questions and if my count was right, you have received approximately 104 answers to all of your questions. Nonetheless you only have 6 upvotes in total. It seems to me that if people are taking their time and making an effort to try to help you with your questions, it is just a matter of courtesy that at least you upvote the answers you receive.2011-05-06
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    Adrian:OK, thanks for your advice, I have never know this tradition before.2011-05-06
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    Well, if an answer is not good, there is no need to upvote it, is there?2011-05-06
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    @Rasmus Of course not, but out of 104 answers, do you think only 6 of them are good? Assuming that the 6 upvotes were given to answers received.2011-05-06
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    @Adri No no, I totally agree that that's too little.2011-05-06
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    @Adrian: Well, I think it's his wish whether to upvote an answer or not.2011-05-19

3 Answers 3

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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 $-\frac{1}{2}$

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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 $$\begin{align*} \lim_{x\to 0} \left(\frac{1}{x}+\frac{\ln(1-x)}{x^2}\right)&= \lim_{x\to 0} \frac{\frac{d}{dx}(x+\ln(1-x))}{\frac{d}{dx}x^2}\\ &= \lim_{x\to 0} \frac{1-\frac{1}{1-x}}{2x}\\ &= \lim_{x\to 0}\frac{\frac{1-x - 1}{1-x}}{2x}\\ &= \lim_{x\to 0} \frac{\frac{x}{x-1}}{2x}\\ &= \lim_{x\to 0}\frac{1}{2(x-1)}\\ &= -\frac{1}{2}. \end{align*}$$

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    How can I break the lines in the equation? I put "\\" but don't work.2011-05-06
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    Someone can help me?2011-05-06
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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 $-\frac1{2}$

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    If this is not so elegant which way is elegant?2011-05-06
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    @Fabian: In general, I feel expanding a function in power series and then computing limits is not the best way to go about. I mean for instance to compute $\lim \frac{1- \cos(\theta)}{\theta^2}$ you could expand $\cos$ in power series or you could write $1 - \cos(\theta) = 2 \sin^2(\theta)$ and proceed. For some reason, I think the second method is elegant as opposed to the first one.2011-05-06
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    When the first few terms of a series expansion are easily available, they provide the best approach. We want to know about the behaviour of the function near $0$. The first few terms of the series give exactly the right information. Note for instance that the second term ($-x/3$) tells us about the "error" when $x$ is near $0$.2011-05-06
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    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