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I've got this limit: $$\displaystyle\lim_{x\to 0} \frac{\ln(1-x)-\sin x}{1-\cos^2 x}$$ and the problem is that it doesn't exist. But I am not very perceptive and I didn't avoid catching in a trap and I started to trying solve this with L'Hôpital's rule. And my question is: are there any ways to notice that given limit doesn't exist in time? If I had been given such a limit on a test, what is the ideal way to solve it?

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Mr. Taylor has a way. – anon Jul 7 '12 at 15:40
Are you watching the movie Mean Girls? See a discussion there. – Did Jul 7 '12 at 16:08
@did, actually I've seen this movie yesterday and this limit interested me :-) thank you very much for the link! – ray Jul 7 '12 at 19:46
up vote 1 down vote accepted

\begin{align} f(x) & = \dfrac{\log(1-x) - \sin(x)}{\sin^2(x)} = \dfrac{\left(-x - \dfrac{x^2}2 - \dfrac{x^3}3 - \cdots \right) - \left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots \right)}{\left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots \right)^2}\\ & = \dfrac{-2x + \mathcal{O}(x^2)}{x^2 + \mathcal{O}(x^3)} = -\dfrac{2+\mathcal{O}(x)}{x+\mathcal{O}(x^2)} \end{align} Hence, $$\lim_{x \to 0^+} f(x) = - \infty$$ $$\lim_{x \to 0^-} f(x) = \infty$$

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+1 because this is the faster method I know of . – Américo Tavares Jul 7 '12 at 19:19

A possible way is to rewrite it as

$$\lim_{x\rightarrow 0}\frac{\ln (1-x)-\sin x}{1-\cos ^{2}x} =\lim_{x\rightarrow 0}\frac{\dfrac{\ln (1-x)}{\sin x}-1}{\sin x}\tag{0}$$

and evaluate by L'Hôpital's rule

$$\lim_{x\rightarrow 0}\frac{\ln (1-x)}{\sin x}=\lim_{x\rightarrow 0}\frac{\frac{-1}{ 1-x }}{\cos x}=-1.\tag{1}$$


$$\lim_{x\rightarrow 0}\frac{\ln (1-x)-\sin x}{1-\cos ^{2}x}=\infty .\tag{2}$$

Note: In view of Marvis' comment I add that this limit is $\infty$ without sign. The side limits depend on the denominator of $(0)$, since its numerator is $-2$ in the limit. For $x>0$ $\sin x>0$ and for $x<0$ $\sin x<0$. Hence

$$\lim_{x\rightarrow 0^+}\frac{\ln (1-x)-\sin x}{1-\cos ^{2}x}=-\infty. \tag{2a}$$

$$\lim_{x\rightarrow 0^-}\frac{\ln (1-x)-\sin x}{1-\cos ^{2}x}=+\infty. \tag{2b}$$

Added: Another way to compute $(1)$ is to write

$$\frac{\ln (1-x)}{\sin x}=\frac{\ln (1-x)}{x}\cdot \frac{x}{\sin x}$$

and use the elementary limits

$$\lim_{x\rightarrow 0}\frac{\ln (1-x)}{x}=-1,$$

$$\lim_{x\rightarrow 0}\frac{x}{\sin x}=1.$$

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-1: since $\lim_{x \to 0^+}$ is $-\infty$ and $\lim_{x \to 0^-}$ is $\infty$ – user17762 Jul 7 '12 at 18:50
@Marvis: I wrote $\infty$ without sign just because of that. – Américo Tavares Jul 7 '12 at 18:52
@Marvis In my notation above $\infty$ does not mean $+\infty$. I added a note with the two side limits. – Américo Tavares Jul 7 '12 at 19:14
Oh Ok. I will remove my down-vote. – user17762 Jul 7 '12 at 19:15
@Marvis Thanks! – Américo Tavares Jul 7 '12 at 19:16

Maclaurin series approach.

$\ln(1-x)\sim -x$, $1-\cos^2 x\sim x^2/2$ and $\sin x\sim x$ for $x \to 0$. Then:

$$\frac{\ln(1-x)-\sin x}{1-\cos^2 x}\sim \frac{-2x}{x^2/2}\to\infty$$

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Can $\sim$ statements be added/subtracted? – anon Jul 7 '12 at 15:46
Sorry, I don't understand. What do you mean? – Mark Jul 7 '12 at 15:48
In this case, I think yes, but in general it might be better to justify by using o/O notation? – Old John Jul 7 '12 at 15:49
It's true that $a\sim b$ and $c\sim d$ imply $a/c\sim b/d$. However, you seem to have combined the $\sim$ statements for $\ln(1-x)$ and $\sin x$ by subtracting the latter from the former. I feel this doesn't work in general, and big-$O$ notation would be justified, as OJ says. – anon Jul 7 '12 at 15:50
Not that it matters, but $1-\cos^2(x)\sim x^2$ – robjohn Jul 7 '12 at 16:03

I am seeing the limit approaching -∞ if x approaches 0 from the positive side, and +∞ if x approaches 0 from the negative side.

I used L'Hôpital's rule one time. The limit is then of the form -2/0. The sign of the limit changes depending if x approaches from the positive side or negative side.

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