Why does the limit $\lim_{x \to 0} \frac{1-\cos^3 x}{x\sin 2x}$ exist? The limit does exist and it is equal to $\frac{3}{4}$. We get it using L'Hospital. 
On the other hand I could write the above limit as:
$$\lim_{x \to 0} \frac{1-\cos^3  x}{x\sin 2x} = \lim_{x \to 0} \frac{1}{x\sin 2x} - \lim_{x \to 0} \frac{\cos^3 x}{x\sin 2x}$$
There is a rule that says, that the limit of difference equals the difference of limits if both the limits in the difference of limits exist.
So in this case the limit $\lim_{x \to 0} \frac{1}{x\sin 2x}$ does not exist as it is not approaching any particular value. So the question is, why does the limit of difference exist if the difference of those two limits doesn't?
 A: The separation of limits into the form
$$\lim_{x\to a}f(x)-g(x)=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)$$
only holds when the limits exist.  In this case, they don't, so we simply aren't allowed to do that, else we would always have
$$\lim_{x\to a}f(x)=\lim_{x\to a}f(x)-\frac1{x-a}+\frac1{x-a}=\underbrace{\lim_{x\to a}f(x)-\frac1{x-a}+\lim_{x\to a}\frac1{x-a}}_{\text{undefined}}$$
A: You have the illustration that two expressions may tend to infinity, yet they approach each other. What's so extraordinary with this? It happens very commonly with asymptotes.
That say, computing the limit is much more illuminating using Taylor's polynomials: you see in depth why the limit is what it is:
First, at order $2$, we know that $\;\cos x=1-\dfrac{x^2}2+o(x^2)$, so
$$1-\cos^3 x=1-\Bigl[\Bigl(1-\dfrac{x^2}2+o(x^2)\Bigr)^3\Bigr]=1-\Bigl[1-\dfrac{3x^2}2+o(x^2)\Bigr]=\dfrac{3x^2}2+o(x^2)$$
On the other hand, $\;x\sin 2x=x\bigl(2x+o(x)\bigr)=2x^2+o(x^2)$, so
$$\frac{1-\cos^3 x}{x\sin 2x}=\frac{\dfrac{3x^2}2+o(x^2)}{2x^2+o(x^2)}=\frac{\dfrac32+o(1)}{2+o(1)}\to\dfrac34.$$
A: If the limits of two functions exist then the limit of their difference (sum, product, quotient provided limit of divisor is non-zero) also exists. This is a standard result proven in most textbooks and you are aware of this result. 
From this result it does not follow that if the limits of two functions do not exist then the limit of their difference does not exist. If $A\Rightarrow B$ then it does not mean that $\neg A\Rightarrow \neg B$.
It is quite possible like in your question that limits of two functions do not exist and yet the limit of their difference exists. In this particular example in your question you may find it difficult to conclude that the limit of difference exists. So consider the simpler case when $\lim_{x\to 0}(1/x\sin x)$ does not exist and yet $$\lim_{x\to 0}\frac{1}{x\sin x} - \frac{1}{x\sin x} =0 $$ in an obvious manner. There may be similar cases where the limit of difference may not exist. Thus $$\lim_{x\to 0}\frac{1}{x\sin x} - \frac{1}{x}$$ does not exist. 
If the hypotheses of a theorem do not hold then it is simply not possible to say anything in general about the conclusions of the theorem. And this is true for the limit theorems which I mentioned at the start of the answer.
The limit in question exists and its existence can be proved via its evaluation as follows
\begin{align}
L&=\lim_{x\to 0}\frac{1-\cos^{3}x}{x\sin 2x}\notag\\
&=\lim_{x\to 0}\frac{1-\cos^{3}x}{1-\cos x} \cdot\frac{1-\cos x} {x\cdot 2x}\cdot\frac{2x}{\sin 2x}\notag \\
&=\lim_{t\to 1}\frac{t^{3}-1}{t-1}\cdot\lim_{x\to 0}\frac{1-\cos x} {2x^{2}}\cdot 1\text{ (putting }t=\cos x) \notag\\
&=\frac{3}{2}\lim_{x\to 0}\frac{1-\cos x} {x^{2}}\notag\\
&=\frac{3}{2}\cdot\frac{1}{2}\notag\\
&=\frac{3}{4}\notag
\end{align}
