Calculate $\lim_{x\to 0}\left(\frac{3\sin(x)-3x \cos(x)}{x^3}\right)^\frac{1}{x}$

Question

This is an exercise from today's exam and I think I did something wrong:

Calculate the limit:

$$\lim_{x\to 0}\left(\frac{3\sin(x)-3x\cos(x)}{x^3}\right)^{1/x}.$$

To tackle this problem I used equivalent infinitesimals: $1-\cos(x) \sim \frac{x^2}{2}$ as $(x\to 0)$ and $\sin(x)\sim x$ as $(x\to 0)$.

Then our limit:

$$\begin{align} \lim_{x\to 0}\left(\frac{3\sin(x)-3x\cos(x)}{x^3}\right)^{1/x}&=\lim_{x\to 0}\left(\frac{3x-3x\cos(x)}{x^3}\right)^{1/x}\\ &=\lim_{x\to 0}\left(\frac{3x(1-\cos(x))}{x^3}\right)^{1/x}\\ &=\lim_{x\to 0}\left(\frac{3x \frac{x^2}{2}}{x^3}\right)^{1/x} \\ &=\lim_{x\to 0}\left(\frac{3}{2}\right)^{1/x} \end{align}$$

Now I split the limit to see what the effect of negative $x$ will be:

$\lim_{x\to 0^+}\left(\frac{3}{2}\right)^{1/x}=+\infty$

$\lim_{x\to 0^-}\left(\frac{3}{2}\right)^{1/x}=0$

I graphed the function and the limit seems to be 1 from both sides.

Answer 1

$$3\sin{x}-3x\cos{x}=3\left(x-\frac{x^3}{6}+\frac{x^5}{120}-...\right)-3x\left(1-\frac{x^2}{2}+\frac{x^4}{24}-...\right)=$$ $$=x^3-\frac{1}{10}x^5+...,$$ which says that your limit is equal to $1$.