Answer
$\dfrac{27}{10}$
Work Step by Step
$\lim\limits_{x \to 0} f(x)=f(0)$
The function will be defined when $c=f(0)$
Thus, $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to 0} \dfrac{9x-3 \sin 3x}{5x^3}=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0} \dfrac{9-9 \cos 3x}{15x^2}=\dfrac{0}{0}$
Again apply L-Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{27 \sin 3x}{30 x}=\dfrac{0}{0}$
Again apply L-Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{81 \cos 3x}{30 }=\dfrac{27}{10}$
Thus, $\lim\limits_{x \to 0} \dfrac{81 \cos (0)}{30 }=\dfrac{27}{10}$
