<s>[INST] $\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\sin x}{x}\lim_{x\to 0} \frac{1}{x^2}\\&= \lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{1}{x^2}\\
&= \lim_{x\to 0} \frac{1}{x^2} -\frac{1}{x^2}\\&=0 \end{align}
But the answer is $\dfrac{1}{2}$ by L'Hopital's Rule.
 [/INST] This is just another way of saying what the others told you.
$$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}
\ne \lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}$$
The theorem is 
IF $\displaystyle \lim_{x\to 0}f(x) = L$ 
and $\displaystyle  \lim_{x\to 0}g(x)=M$, where $M, N \in \mathbb R$,
THEN $\displaystyle \lim_{x\to 0}(f(x)-g(x))=L-M$
But, since $\displaystyle \lim_{x\to 0} \frac{\tan x}{x^3} = \lim_{x\to 0} \frac{\sin x}{x^3} = \infty$, then the theorem does not apply.
This limit can be evaluated without resorting to L'Hospital.
\begin{align}
   \frac{\tan x - \sin x}{x^3}
   &= \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \\
   &= \frac{\sin x - \sin x \cos x}{x^3 \cos x} \\
   &= \frac{1}{\cos x} \cdot\frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} \\
   &= \frac{1}{\cos x} \cdot\frac{\sin x}{x} 
      \cdot \frac{2\sin^2(\frac 12x)}{x^2} \\
      &= \frac{1}{\cos x} \cdot\frac{\sin x}{x} 
      \cdot \frac 12 \cdot \left(\frac{\sin \frac x2}{\frac x2}\right)^2 \\
\end{align}
which approaches $\dfrac 12$ as $x$ approaches $0$.
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