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?

You were on the right track with the second equality, all that was left was to multiply numerator and denominator by a "conjugate"
$$\frac{\sin x(\sec x - 1)}{x^4}\cdot\frac{\sec x+1}{\sec x + 1} = \frac{\sin x}{x}\cdot \frac{\tan^2x}{x^2} \cdot\frac{1}{x} \cdot\frac{1}{\sec x + 1}$$
The limit to $0$ of $xf(x)$ is
$$ 1\cdot 1^2 \cdot \frac{1}{1+1} = \frac{1}{2}$$
Thus the limit of $f$ to $0$ does not exist.
