<s>[INST] Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{x(1 - 0.5\cos x) - 0.5\sin x}}{{{x^3}}}$ For evaluating the limit $\lim\limits_{x \to 0} \frac{x(1 - 0.5\cos x) - 0.5\sin x}{x^3}$, I proceeded as follows:
$$\lim_{x \to 0} \left(\frac{x(1 - 0.5\cos x)}{x^3} - \frac{0.5}{x^2}\left(\frac{\sin x}{x}\right)\right)$$
Using the fact that $\lim\limits_{x \to 0} \frac{\sin x}{x}=1$ and $\lim\limits_{x \to a} \{ f(x) - g(x)\}  = \lim\limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$, I got 
$$\lim\limits_{x \to 0} \frac{1 - 0.5\cos x}{x^2}  - \lim\limits_{x \to 0} \frac{0.5}{x^2}$$
$$ = \lim_{x \to 0} \left(\frac{1 - 0.5\cos x - 0.5}{x^2} \right)$$
Now, after applying L'Hopital's Rule I got the final answer as $0.25$. However, on evaluating the original limit using Mathematica, I got the answer as $\frac{1}{3}$.
Can someone please tell me where am I going wrong. 
Somehow, I believe that you cannot use the fact that $\lim\limits_{x \to a} \{ f(x) - g(x)\}  = \lim \limits_{x \to a} f(x) - \lim\limits_{x \to a} g(x)$ in case of indeterminate forms.
Thanks in advance!
 [/INST] Apply iteratedly l'Hôpital Rule to $$\frac{x(2-\cos x)-\sin x}{2x^3}$$ We get successively  $$\frac{2-2\cos x+x\sin x}{6x^2}$$  $$\frac{3\sin x+x\cos x}{12x}$$  $$\frac{3\cos x+\cos x-x\sin x}{12}$$
Thus the searched limit is the limit of $$\frac{4\cos x-x\sin x}{12}$$ when $x$ tends to $0$ which is clearly $\frac{4}{12}=\frac 13$
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