$$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
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.
On
I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ \lim\limits_{x \to 0}\big( f(x) - g(x)\big)$ is not always equal to $ \lim\limits_{x \to 0} f(x) - \lim\limits_{x \to 0} g(x)$.
On
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 $0$.
On
Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $\sin$ & $\tan$ as Taylor series - each has first term in $\theta^1$ with coefficient 1, so in the difference it drops out. If you plot $\sin\theta-\tan\theta$ it looks like a cubic at the origin. Then if you divide that series by $\theta^3$, & you get a series with an initial term in $\theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $\theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $\theta\rightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.
To actually get the answer immediately you just subtract the coefficient for $\theta^3$ in the series for $\sin\theta$ from that in that for $\tan\theta$, & you get 1/3 - -1/6 = 1/2.
On
Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.
We have
$$\lim_{x \to 0} \tan x= \frac{x}{1} +\frac{x^{3}}{3} +\frac{2x^{5}}{15} + . . .$$
$$\lim_{x \to 0} \sin x= \frac{x}{1} - \frac{x^{3}}{6} +\frac{x^{5}}{120} + . . .$$
Therefore expression turns to,
$$\lim_{x \to 0} \frac{\frac{x}{1} +\frac{x^{3}}{3} +\frac{2x^{5}}{15} + . . . - (\frac{x}{1} - \frac{x^{3}}{6} +\frac{x^{5}}{120} + . . .)}{x^{3}}$$
Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $\frac{1}{3}$ and $\frac{1}{6}$ which is $\color{red} {\frac{1}{2}}$
Your problem arises from the fact that you used $\color{red}{\lim_\limits{x \to 0} \frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $\color{red}{\infty-\infty}$...
Only split an initial limit into a product if the individual limits are defined.