Evaluate $\lim_{x\to 0} (\ln(1-x)-\sin x)/(1-\cos^2 x)$ 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?
 A: 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
$$\lim_{x\rightarrow 0}\frac{\ln (1-x)}{\sin x}=\lim_{x\rightarrow 0}\frac{\frac{-1}{ 1-x }}{\cos x}=-1.\tag{1}$$
Consequently,
$$\lim_{x\rightarrow 0}\frac{\ln (1-x)-\sin x}{1-\cos ^{2}x}=\infty .\tag{2}$$
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
$$\lim_{x\rightarrow 0}\frac{\ln (1-x)}{x}=-1,$$
$$\lim_{x\rightarrow 0}\frac{x}{\sin x}=1.$$
A: \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, $$\lim_{x \to 0^+} f(x) = - \infty$$ $$\lim_{x \to 0^-} f(x) = \infty$$
A: 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:
$$\frac{\ln(1-x)-\sin x}{1-\cos^2 x}\sim \frac{-2x}{x^2/2}\to\infty$$
A: 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.
