$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}$ Is this solution correct? Just want to double check whether all my operations are legal. The result seems to be correct but I want to make sure I didn't make any mistakes.
$\require{cancel}$
$$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{\sin x}{e^x - \cos x}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{x}{e^x - \cos x}\cdot\cancelto{1}{\frac{\sin x}{x}}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2+x}{e^x - 1 + 1 - \cos x}=$$
$$=\lim_{x \to 0} (\frac {1-\cos x}{x^{11} - 3x^2 + x} + \frac {e^x-1}{x^{11}-3x^2+x})^{-1}=$$
$$=\lim_{x \to 0} (\frac {1-\cos x}{x^2\cancelto{-\infty}{(x^9-3-\frac {1}{x})}} + \frac {e^x-1}{x\cancelto{1}{(x^{10}-3x+1)}})^{-1}=$$
$$ =\lim_{x \to 0} (\cancelto{\frac{1}{2}}{\frac{1-\cos x}{x^2}}\cdot\cancelto{0}{\frac{1}{-\infty}} + \cancelto{1}{\frac {e^x-1}{x}})^{-1}=$$
$$=\lim_{x \to 0} (\frac {1}{2} \cdot0+ 1)^{-1}=$$
$$=(0+1)^{-1} = 1$$
Is this correct?
 A: The final result is correct, but in my opinion  your approach is too complicated and you should avoid the "inverse of the inverse" part. More simply, by using the same standard limits as $x\to 0$, 
$$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{\overbrace{x^{10}-3x}^{\to 0}+\overbrace{\frac{\sin x}{x}}^{\to 1}}{\underbrace{\frac{e^x-1}{x}}_{\to 1} + \underbrace{\frac{1- \cos x}{x^2}}_{\to 1/2}\cdot\underbrace{x}_{\to 0}}\to 1.$$
If you are comfortable with the little-o notation, there is a shorter way: as $x\to 0$,
$$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{x^{11}-3x^2+(x+o(x))}{(1+x+o(x)) - (1-o(x))}=\frac{x+o(x)}{x+o(x)}\to 1.$$
A: L'Hopital's rule also works as the limit is in the form $\frac{0}{0}$:
$$\lim_{x \to 0}  \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}
=\lim_{x \to 0}  \frac{11x^{10}-6x+\cos x}{e^x + \sin x}
= \frac{0-0+1}{1 + 0}=1$$
A: The result is correct but we can simplify some step as follows
$$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{x^{11}-3x^2}{e^x - \cos x}+\frac{\sin x}{e^x - \cos x}$$
and since the first term tends to $0$
$$\frac{x^{11}-3x^2}{e^x - \cos x}=\frac{x^{10}-3x}{\frac{e^x - 1}x+x\frac{1-\cos x}{x^2}}\to 0$$
we obtain
$$\frac{\sin x}{e^x - \cos x}=\frac{\frac{\sin x}x}{\frac{e^x - 1}x+x\frac{1-\cos x}{x^2}}\to 1$$
