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?
 A: Hints: $\quad\ln(ab)=\ln a+\ln b,\qquad\ln e^x=x,\qquad\cos x=\sqrt{1-\sin^2x},\qquad\ln a^b=b\ln a,$
$\ln(1+t)\simeq t\quad$ when $\quad t\simeq0,\quad$ and $\quad\lim_{x\to0}\dfrac{\sin x}x=1$.
A: 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 $L = 1/2$.
A: 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 $\sin x/x$.
A: yes it is ${\frac {1}{2}}-{\frac {1}{6}}x+{\frac {1}{6}}{x}^{2}-{\frac {7}{360}}
{x}^{3}+O \left( {x}^{4} \right) 
$
A: Hint: Just use Taylor expension, $\sin x=x+o(x^{2}),\;\cos x=1-\frac{x^2}{2!}+o(x^2),\;$
$\ln(1+x)=x-\frac{x^2}{2}+o(x^2).$
Then
$\ln(e^x\cos x)=x+\ln\cos x=x-\frac{x^2}{2}+o(x^2)$. Hence
$\lim_{x\to 0} \frac{\sin{x}-\ln({\text{e}^{x}}\cos{x})}{x\sin{x}} =
\lim_{x\to0} \frac{x-(x-\frac{x^2}{2})+o(x^2)}{x^2} =\frac{1}{2} .$
