Answer
$3$
Work Step by Step
Here, we have $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0} \dfrac{3 \sin x+(3x+1)(cos x)-1}{\sin x+x\cos x}=\dfrac{0}{0}$
Now, again use L-Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{6 \cos x-(3x+1)(\sin x)}{2\cos x-x \sin x}=\dfrac{6-0}{2-0}$
or, $\dfrac{6}{2}=3$
