# How to calculate $\lim\limits_{x \to 0}\frac1{x} + \frac{\ln(1-x)}{x^2}$?

How to calculate the following limit?

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

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Well, if an answer is not good, there is no need to upvote it, is there? – Rasmus May 6 '11 at 17:49
@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. – Adrián Barquero May 6 '11 at 18:07
@Adri No no, I totally agree that that's too little. – Rasmus May 6 '11 at 18:09

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? – Fabian May 6 '11 at 17:38
@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. – user17762 May 6 '11 at 17:40
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 $0$. – André Nicolas May 6 '11 at 18:27
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. :-) – Did May 7 '11 at 9:04

You have \begin{eqnarray*} \lim_{x\to 0} \frac{1}{x}+\frac{\ln(1-x)}{x^2} &=& \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} \frac{1}{x}+\frac{\ln(1-x)}{x^2}&= \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. – leo May 6 '11 at 18:09
Someone can help me? – leo May 6 '11 at 18:17
Thank You @J.M. – leo May 6 '11 at 18:20

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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