We are given, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x + 3x}}{{2x + \tan 3x}}$.

Multiplying and dividing the numerator by $2x$ then,

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x + 3x}}{{2x + \tan 3x}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x(\sin 2x)/2x + 3x}}{{2x + 3x(\tan 3x)/3x}}$

$ = \frac{{\mathop {\lim }\limits_{x \to 0} 2x \cdot \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin 2x}}{{2x}}} \right] + \mathop {\lim }\limits_{x \to 0} 3x}}{{\mathop {\lim }\limits_{x \to 0} 2x + \mathop {\lim }\limits_{x \to 0} 3x\mathop {\lim }\limits_{X \to 0} \left[ {\frac{{\tan 3x}}{{3x}}} \right]}}$

Since, $\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\tan 3x}}{{3x}}} \right]$ and $\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin 2x}}{{2x}}} \right]$ both are equal to $1$.

So,

$\frac{{\mathop {\lim }\limits_{x \to 0} 2x \cdot \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin 2x}}{{2x}}} \right] + \mathop {\lim }\limits_{x \to 0} 3x}}{{\mathop {\lim }\limits_{x \to 0} 2x + \mathop {\lim }\limits_{x \to 0} 3x \cdot \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\tan 3x}}{{3x}}} \right]}} = \frac{{\mathop {\lim }\limits_{x \to 0} 2x \cdot 1 + \mathop {\lim }\limits_{x \to 0} 3x}}{{\mathop {\lim }\limits_{x \to 0} 2x + \mathop {\lim }\limits_{x \to 0} 3x \cdot 1}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2x + 3x}}{{2x + 3x}}$

$ = \mathop {\lim }\limits_{x \to 0} 1$

$ = 1$

Therefore, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x + 3x}}{{2x + \tan 3x}} = 1$.