How can I calculate $\lim_{x \to 0}\left(\frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}}\right)$ limit without using L'Hopital's rule? $$\lim_{x \to 0}\left(\frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}}\right)$$
Can this limit be calculated without using L'Hopital's rule?

Clearly $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sin x - \log(e^{x}\cos x)}{x\sin x}\\
&= \lim_{x \to 0}\frac{\sin x - x - \log \cos x}{x^{2}}\cdot\frac{x}{\sin x}\\
&= \lim_{x \to 0}\frac{\sin x - x - \log \cos x}{x^{2}}\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{2}} - \frac{\log \cos x}{x^{2}}\\
&= A - B\end{aligned}$$ Next we can see that $$A = 0$$ from here. And we have $$\begin{aligned}B &= \lim_{x \to 0}\frac{\log\cos x}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log\cos x}{\cos x - 1}\cdot\frac{\cos x  - 1}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log\cos x}{\cos x - 1}\cdot\lim_{x\to 0}\frac{\cos x  - 1}{x^{2}}\\
&= \lim_{t \to 0}\frac{\log(1 + t)}{t}\cdot\lim_{x\to 0}\frac{\cos x  - 1}{x^{2}}\text{ (putting } t = \cos x - 1)\\
&= \lim_{x \to 0}\frac{\cos x  - 1}{x^{2}}\\
&= \lim_{x \to 0}\frac{\cos^{2} x - 1}{x^{2}(1 + \cos x)}\\
&= -\frac{1}{2}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}} = -\frac{1}{2}\end{aligned}$$ It follows that $L = 1/2$.
