To solve the limit $\lim _{x\to 0}\frac{\sin \left(4x\right)\cdot \tan ^2\left(3x\right)\:+\:6x^2}{2x^2+\sin \left(3x\right)\cdot \cos \left(2x\right)}$ using trigonometric identities, let's analyze each component of the expression using small-angle approximations.

Step 1: Small-Angle Approximations

For $x$ close to 0, we can use the following approximations:

$\sin(y) \approx y \quad \text{and} \quad \tan(y) \approx y \quad \text{for small } y$

Substitute these approximations into the expression.

Step 2: Simplify the Numerator

For the numerator $\sin(4x) \cdot \tan^2(3x) + 6x^2$, we have:

$\sin(4x) \approx 4x$

$\tan^2(3x) \approx (3x)^2 = 9x^2$

Thus, the numerator simplifies to:

$\sin(4x) \cdot \tan^2(3x) + 6x^2 \approx 4x \cdot 9x^2 + 6x^2 = 36x^3 + 6x^2$

Step 3: Simplify the Denominator

For the denominator $2x^2 + \sin(3x) \cdot \cos(2x)$, we have:

$\sin(3x) \approx 3x$

\cos(2x) \approx 1 \quad \text{(since \cos(y) \approx 1 - \frac{y^2}{2}$and the higher order terms vanish as$y \to 0)}

Thus, the denominator simplifies to:

$2x^2 + \sin(3x) \cdot \cos(2x) \approx 2x^2 + 3x \cdot 1 = 2x^2 + 3x$

Step 4: Express