Interpreting your expression as
[tex] \dfrac{3(1+2x)}{\sin(x)} [/tex]
when [tex] x [/tex] approaches zero, the numerator approaches 3:
[tex] 3(1+2x) \to 3(1+2\cdot 0) = 3(1+0) = 3\cdot 1 = 3 [/tex]
The denominator approaches 0, because [tex] \sin(0)=0 [/tex]
Moreover, we have
[tex] \displaystyle \lim_{x\to 0^-} \sin(x) = 0^-,\quad \displaystyle \lim_{x\to 0^+} \sin(x) = 0^+ [/tex]
So, the limit does not exist, because left and right limits are different:
[tex] \displaystyle \lim_{x\to 0^-} \dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^-} = -\infty,\quad \displaystyle \lim_{x\to 0^+}\dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^+} = +\infty [/tex]
