Let's analyze the given limit problem step by step:

The problem is to evaluate the limit: $\lim_{x \to 0} \frac{\tan x^2}{\sin 2x \cdot (x^2 + 3x)}$

Step-by-step solution:

1. Rewrite the numerator using small-angle approximations:

   • As $x \to 0$, $\tan x^2 \approx x^2$.
   • Thus, the limit becomes: $\lim_{x \to 0} \frac{x^2}{\sin 2x \cdot (x^2 + 3x)}$

2. Simplify the denominator:

   • Note that $\sin 2x \approx 2x$ when $x \to 0$.
   • Thus, the limit can be rewritten as: $\lim_{x \to 0} \frac{x^2}{2x (x^2 + 3x)}$

3. Factor and simplify:

   • Simplify the expression inside the limit: $\frac{x^2}{2x (x^2 + 3x)} = \frac{x^2}{2x^3 + 6x^2} = \frac{x^2}{2x^2 (x + 3)} = \frac{1}{2(x + 3)}$

4. Evaluate the limit:

   • As $x \to 0$: $\lim_{x \to 0} \frac{1}{2(x + 3)} = \frac{1}{2(0 + 3)} = \frac{1}{6}$

Therefore, the value of the limit is: $\boxed{\frac{1}{6}}$

Would you like more details on any specific part of this solution? Here are some questions you might find useful:

1. How do small-angle approximations work?
2. Can you explain more about simplifying limits?
3. What are common limit techniques in calculus?
4. How do you handle indeterminate forms?
5. Can you show another example of evaluating a limit?

Tip: Always remember to check if the function can be simplified using algebraic manipulations before applying limit properties or theorems.