Applying L'Hopital's rule for $\lim_{x\to 0}\frac{e^x\sin x -x(1-x)}{x^3}$

I need to find

$$\lim_{x\to 0}\frac{e^x\sin x -x(1-x)}{x^3}$$.

If I apply L'Hôpital's rule two times I get

$$\lim_{x\to 0^{+}}\frac{e^x\cos x +1}{3x} = \infty$$

$$\lim_{x\to 0^{-}}\frac{e^x\cos x +1}{3x} = -\infty$$

can I assume because I didn't get a limit, the original function doesn't have a limit or not? Because from what I read here, I can't assume anything. If I can't, is there another way to get the limit?

Actually,$$\lim_{x\to0^+}\frac{e^x\cos(x)+1}{3x}=\infty,$$and it follows from L'Hopital's Rule that$$\lim_{x\to0^+}\frac{e^x\sin(x)-x(1-x)}{x^3}=\infty.$$By the same argument,$$\lim_{x\to0^-}\frac{e^x\sin(x)-x(1-x)}{x^3}=-\infty.$$So, the limit$$\lim_{x\to0}\frac{e^x\sin(x)-x(1-x)}{x^3}$$doesn't exist indeed.

As an alternative way, we can avoid l'Hopital's rule noting that

$$\frac{e^x\sin x -x(1-x)}{x^3}=\frac{e^x\sin x -\sin x+\sin x-x(1-x)}{x^3}=$$

$$=\frac{e^x\sin x -\sin x}{x^3}+\frac{\sin x-x(1-x)}{x^3}=$$

$$=\frac 1 x \frac{\sin x}x\frac{e^x -1}{x}+\frac{\sin x-x+x^2}{x^3}=$$

$$=\frac 1 x \frac{\sin x}x\frac{e^x -1}{x}+\frac{\sin x-x}{x^3}+\frac{1}{x}$$

with $\frac{\sin x}x \to 1$, $\frac{e^x -1}{x}\to 1$, $\frac{\sin x-x}{x^3}\to -\frac 16$ but limit for $\frac 1 x$ doesn't exist.