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!
 A: Use  Taylor's development at order $3$:


*

*$\cos x= 1-\dfrac{x^2}2+o(x^2)$, hence $\;x\bigl(1-\frac12\cos x\big)=\dfrac x2+\dfrac{x^3}4+o(x^3)$,

*$\sin x=x-\dfrac{x^3}6+o(x^3)$.


Thus the numerator is
$$\dfrac x2+\dfrac{x^3}4-\frac x2+\dfrac{x^3}{12}+o(x^3)=\dfrac{x^3}3+o(x^3)$$
and finally
$$\frac{x\bigl(1-\frac12\cos x\big)-\frac12\sin x}{x^3}=\frac{\dfrac{x^3}3+o(x^3)}{x^3}=\frac13+o(1)\to\frac13.$$
A: 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 $\frac{4}{12}=\frac 13$
A: If you want to compute this limit by making use of known basic limits then
use the following
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}=-\frac{1}{6},\ \ \ \ and\ \ \ \
\ \ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}
\end{equation*}
as follows
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{x(1-\frac{1}{2}\cos x)-\frac{1}{2}\sin x}{x^{3}}
&=&\lim_{x\rightarrow 0}\frac{1}{2}\frac{2x(1-\frac{1}{2}\cos x)-\sin x}{%
x^{3}} \\
&=&\lim_{x\rightarrow 0}\frac{1}{2}\frac{x+x-x\cos x-\sin x}{x^{3}} \\
&=&\lim_{x\rightarrow 0}\frac{1}{2}\left( \frac{x-\sin x}{x^{3}}+\frac{%
1-\cos x}{x^{2}}\right)  \\
&=&\frac{1}{2}\left( \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}%
+\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}\right)  \\
&=&\frac{1}{2}\left( -\frac{1}{6}+\frac{1}{2}\right) =\frac{1}{6}.
\end{eqnarray*}
Warning: When we write
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}+\frac{1-\cos x}{x^{2}}%
\right) =\lim_{x\rightarrow 0}\left( \frac{x-\sin x}{x^{3}}\right)
+\lim_{x\rightarrow 0}\left( \frac{1-\cos x}{x^{2}}\right) 
\end{equation*}
it is because we $\textbf{previously}$ know that the limits of the RHS exist
as real numbers.
