Using Taylor expansion to find $\lim_{x \rightarrow 0} \frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}$ 
Use Taylor expansion to find $\displaystyle \lim_{x \rightarrow 0} \frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}$

I know that:
$\exp(2x) = 1 + (2x) + \frac{(2x)^2}{2!}+ O(x^3)$
$\ln(1-x) = 1-(-x) + (-x)^2 + O(x^3)$
$\sin(x) = x + O(x^3)$
$\cos(x) = 1 - \frac{x^2}{2!} + O(x^3)$
Substituting this in yields:
$\displaystyle \lim_{x \rightarrow 0} \frac{1+2x+2x^2+O(x^3) -1 - x - x^2 + O(x^3) -x + O(x^3)}{-\frac{x^2}{2!} + O(x^3)} = \lim_{x \rightarrow 0} \frac{x^2 + O(x^3)}{-\frac{1}{2}x^2 + O(x^3)} = \lim_{x \rightarrow 0} \frac{x^2[1+O(x)]}{x^2[-1/2+O(x)]} = -2$

EDIT: Progressing from $\displaystyle \lim_{x \rightarrow 0} \frac{1+2x+\frac{5}{2}x^2+O(x^3)}{-\frac{x^2}{2}+O(x^3)}$ we have,
$\displaystyle \lim_{x \rightarrow 0} \frac{1 + 5/2x^2 +O(x^3)}{-1/2x^2 + O(x^3)} = \lim_{x \rightarrow 0} \frac{x^2(1/x^2 + 5/2 + O(x)}{x^2(-1/2 + O(x))} = \lim_{x \rightarrow 0} \frac{-1/x^2 - 5/2 + O(x)}{1/2+O(x)} = \frac{\lim_{x \rightarrow 0} -1/x^2 - 5/2 + O(x) }{\lim_{x \rightarrow 0} 1/2 + O(x)} = \frac{-\infty}{1/2} = -\infty$
Is this method rigorously correct? 
 A: Your Taylor expansion of $\ln(1-x)$ is wrong
It should be $\displaystyle \ln(1-x)=-x-\frac{x^2}{2}+O(x^3)$
Therefore, $$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}=\frac{1+2x+\frac{5}{2}x^2+O(x^3)}{\frac{-x^2}{2}+O(x^3)}$$
The last expression goes to $-\infty$

Further explanation of this last fact:
$$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}=\frac{1+2x+\frac{5}{2}x^2+O(x^3)}{\frac{-x^2}{2}+O(x^{\color{red}{4}})}=\frac{1+2x+O(x^2)}{\frac{-x^2}{2}(1+O(x^2))}$$
Since $\displaystyle\frac{1}{1+\color{red}{O(x^2)}}=1-\color{red}{O(x^2)}+O(\color{red}{O(x^2)})=1+O(x^2)$,
$$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}=\frac{\left(1+2x+O(x^2)\right)\left(1+O(x^2)\right)}{\frac{-x^2}{2}}$$
Expanding the numerator, you get $\left(1+2x+O(x^2)\right)\left(1+O(x^2)\right)=1+2x+O(x^2)$
So : $\displaystyle \frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}=\frac{1+2x+O(x^2)}{\frac{-x^2}{2}}$
Hence, 
$$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}= \frac{-2}{x^2}-\frac{4}{x}+O(1)$$
But, at $0$, $\displaystyle \frac{1}{x}=o(\frac{1}{x^2})$
Finally $$\frac{\exp(2x)-\ln(1-x)-\sin(x)}{\cos(x)-1}= \frac{-2}{x^2}+o(\frac{1}{x^2})+O(1)$$
The last quantity goes to $-\infty$
