$$\lim_{x \rightarrow \infty}\frac{(2x^{2}-3x+5)(3x-1)^{\frac{x}{2}}}{(3x^{2}+5x+4)\sqrt{(3x+2)^{x}}}$$ is equals to :

We need to evaluate:

$$\lim_{x \rightarrow \infty}\frac{(2x^{2}-3x+5)(3x-1)^{\frac{x}{2}}}{(3x^{2}+5x+4)\sqrt{(3x+2)^{x}}}$$

Let us split this into two parts: a rational part and an exponential part.

Part 1: The rational fraction

$$\frac{2x^{2} - 3x + 5}{3x^{2} + 5x + 4}$$

Dividing numerator and denominator by $$x^{2}$$:

$$\frac{2 - \frac{3}{x} + \frac{5}{x^{2}}}{3 + \frac{5}{x} + \frac{4}{x^{2}}}$$

As $$x \rightarrow \infty$$, the terms with $$x$$ in the denominator vanish, so this approaches:

$$\frac{2}{3}$$

Part 2: The exponential fraction

We simplify the exponential part. Note that $$\sqrt{(3x+2)^{x}} = (3x+2)^{x/2}$$. So the exponential part becomes:

$$\frac{(3x-1)^{x/2}}{(3x+2)^{x/2}} = \left(\frac{3x-1}{3x+2}\right)^{x/2}$$

Now we write:

$$\frac{3x - 1}{3x + 2} = 1 - \frac{3}{3x + 2}$$

So we need:

$$\lim_{x \rightarrow \infty} \left(1 - \frac{3}{3x + 2}\right)^{x/2}$$

This is of the form $$1^{\infty}$$. We take the natural logarithm of the expression. Let:

$$L = \lim_{x \rightarrow \infty} \frac{x}{2} \cdot \ln\left(1 - \frac{3}{3x + 2}\right)$$

For large $$x$$, using the approximation $$\ln(1 + u) \approx u$$ when $$u \rightarrow 0$$:

$$\ln\left(1 - \frac{3}{3x + 2}\right) \approx -\frac{3}{3x + 2}$$

Therefore:

$$L = \lim_{x \rightarrow \infty} \frac{x}{2} \cdot \left(-\frac{3}{3x + 2}\right)$$

$$L = \lim_{x \rightarrow \infty} \frac{-3x}{2(3x + 2)}$$

$$L = \lim_{x \rightarrow \infty} \frac{-3x}{6x + 4}$$

Dividing numerator and denominator by $$x$$:

$$L = \lim_{x \rightarrow \infty} \frac{-3}{6 + \frac{4}{x}} = \frac{-3}{6} = -\frac{1}{2}$$

So the exponential part equals $$e^{L} = e^{-1/2} = \frac{1}{\sqrt{e}}$$.

Combining both parts:

$$\lim_{x \rightarrow \infty}\frac{(2x^{2}-3x+5)(3x-1)^{x/2}}{(3x^{2}+5x+4)(3x+2)^{x/2}} = \frac{2}{3} \cdot \frac{1}{\sqrt{e}} = \frac{2}{3\sqrt{e}}$$

Hence, the correct answer is Option C.