calculus - How can I calculate this 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?

Answer

Rewriting the numerator as $\sin x-x-\ln\cos x$, notice that $\sin x-x$ is an odd function; as the denominator is even, the ratio vanishes at $x=0$ and these terms can be ignored.

Then

$$\lim_{x\to0}-\frac{\ln\cos x}{x\sin x}=-\lim_{x\to0}\frac{\frac12\ln(1-\sin^2x)}{\sin^2x}\frac{\sin x}x=-\frac12\lim_{t\to0^+}\frac{\ln(1-t)}t=-\frac12\lim_{t\to0^+}\ln(1-t)^{1/t}\\=-\frac12\ln\left(\lim_{t\to0^+}(1-t)^{1/t}\right)=-\frac12\ln e^{-1}=\frac12.$$

UPDATE:

We need to show that the first limit exists. Without being allowed to use derivatives, we need some property of the trigonometric functions, and we admit $\sin x\le x\le \tan x$, so that

$$1\ge\frac{\sin x}x\ge\cos x\ge\cos^2x,$$
and from there

$$0\ge\frac{\sin x-x}x\ge\cos^2x-1,$$
$$0\ge\frac{\sin x-x}{x\sin x}\ge-\sin x.$$
As expected, the limit is $0$. As a byproduct, the same relations establish the limit of $\sin x/x$.