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?

\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, $$\lim_{x \to 0^+} f(x) = - \infty$$ $$\lim_{x \to 0^-} f(x) = \infty$$