$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}$ 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?

L'Hopital's rule also works as the limit is in the form $\frac{0}{0}$:
$$\lim_{x \to 0}  \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}
=\lim_{x \to 0}  \frac{11x^{10}-6x+\cos x}{e^x + \sin x}
= \frac{0-0+1}{1 + 0}=1$$
