Answer
$3$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{3 \sin x+(3x+1)(cos x)-1}{\sin x+x\cos x}=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{6 \cos-(3x+1)(\sin x)}{2\cos x-x \sin x}=\dfrac{6-0}{2-0}=3$
