Find the trigonometric limit Find the limit $$\lim_{x \to 0} \frac{\tan x-\sin x}{x^4} $$
Here is my solution:
$$\lim_{x \to 0} \frac{\tan x-\sin x}{x^4} = \lim_{x \to 0}\frac{\sin x(\frac{1}{\cos x}-1)}{x^4}=\lim_{x \to 0}\frac{\sin x}{x^4}*(\frac{1}{\cos x}-1)=\lim_{x \to 0}\frac{\sin x}{x^4}*(0)$$
$$ \lim_{x \to 0}\frac{\sin x}{x^4}$$
$$ \to \lim_{x \to 0+0}\frac{\sin(0+0)}{(0+0)^4}=\frac{0}{0} \space AND \space \lim_{x \to 0-0}\frac{\sin(0-0)}{(0-0)^4}=\lim_{x \to 0-0}\frac{-\sin(0)}{(0)^4}=-\frac{0}{0}, Hence \space the \space limit \space DNE $$
So, is that an appropriate way? I have troubles with the limits that have no solution (do not exist). I can easily find the limit with L'Hôpital's Rule, but then it turns out that the limit one-sided and do not exist. Is there any universal method for that type of limits?

For $x\approx0$, we know that $\sin x\approx x-\frac{x^3}{6}$ and $\tan x \approx x+\frac{x^3}{3}$.* Hence,
\begin{align}
\frac{\tan x-\sin x}{x^4} &\approx \frac{x+\frac{x^3}{3}-x+\frac{x^3}{6}}{x^4} \\[5pt]
&= \frac{\frac{x^3}{2}}{x^4} \\[5pt]
&= \frac{1}{2x} \, .
\end{align}
We thus conclude that the limit from the right is $\infty$, and the limit from the left is $-\infty$.

*(The higher-order terms of the Taylor series of $\sin x$ and $\tan x$ can safely be ignored because they contain powers of $x$ which are larger than $4$.)
