<s>[INST] Calculate $\lim_{x\to 0}\left(\frac{3\sin(x)-3x \cos(x)}{x^3}\right)^\frac{1}{x}$ 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.
 [/INST] $$3\sin{x}-3x\cos{x}=3\left(x-\frac{x^3}{6}+\frac{x^5}{120}-...\right)-3x\left(1-\frac{x^2}{2}+\frac{x^4}{24}-...\right)=$$
$$=x^3-\frac{1}{10}x^5+...,$$
which says that your limit is equal to $1$.
</s>