\section*{Exercise 2 - V01102}

Determine the limit of the function:

\[
\lim _{x \rightarrow 0} \frac{(1-x)^{6} \sin 3 x-\sin 3 x}{3 x^{2}}
\]
}

For \(x=0\) an indeterminate form \(\frac{0}{0}\) arises. According to l'Hôpital's rule, the limit in this case is given by

\[
\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{g^{\prime}(x)}
\]

Twice differentiating the numerator and the denominator (not using the quotient rule!) yields:

\[
\begin{aligned}
f^{\prime \prime}(x) & =36\left(x^{5}-5 x^{4}+10 x^{3}-10 x^{2}+5 x-1\right) \cos 3 x- \\
& -3\left(3 x^{6}-18 x^{5}+35 x^{4}-20 x^{3}-15 x^{2}+22 x-10\right) \sin 3 x \\
g^{\prime \prime}(x) & =6
\end{aligned}
\]

Thus, the value

\[
\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{g^{\prime \prime}(x)}=-6
\]

is the limit of the initial expression.