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The given limit is:

\[
\lim_{{x \to +\infty}} \frac{3x\sqrt{x^2 + x - 3x^2 + x}}{5\sqrt{x^2 + 3}}
\]

### Solution:

1. **Simplify the expressions inside the square roots**:
   - The numerator inside the square root:
     \[
     x^2 + x - 3x^2 + x = -2x^2 + 2x
     \]
   - So, the numerator becomes:
     \[
     3x\sqrt{-2x^2 + 2x}
     \]
   - The denominator is:
     \[
     5\sqrt{x^2 + 3}
     \]

2. **Factor out the highest power of \(x\)**:
   - In the numerator, factor \(x^2\) out of the square root:
     \[
     \sqrt{-2x^2 + 2x} = x\sqrt{-2 + \frac{2}{x}}
     \]
   - The numerator simplifies to:
     \[
     3x \cdot x \sqrt{-2 + \frac{2}{x}} = 3x^2 \sqrt{-2 + \frac{2}{x}}
     \]
   - In the denominator, factor \(x^2\) from the square root:
     \[
     \sqrt{x^2 + 3} = x\sqrt{1 + \frac{3}{x^2}}
     \]
   - The denominator simplifies to:
     \[
     5x\sqrt{1 + \frac{3}{x^2}}
     \]

3. **Simplify the limit as \(x \to \infty\)**:
   - As \(x \to \infty\), the terms \(\frac{2}{x}\) and \(\frac{3}{x^2}\) approach 0, so:
     \[
     \sqrt{-2 + \frac{2}{x}} \to \sqrt{-2}
     \]
     and
     \[
     \sqrt{1 + \frac{3}{x^2}} \to \sqrt{1} = 1
     \]
   - Therefore, the expression simplifies to:
     \[
     \lim_{{x \to +\infty}} \frac{3x^2 \cdot \sqrt{-2}}{5x \cdot 1} = \frac{3x^2 \cdot i\sqrt{2}}{5x}
     \]

4. **Final simplification**:
   - Cancel out \(x\) terms:
     \[
     \frac{3x \cdot i\sqrt{2}}{5}
     \]
   - As \(x \to \infty\), the limit does not exist in the real numbers, since the square root of a negative number involves imaginary numbers. Therefore, the limit approaches an imaginary value:
     \[
     \lim_{{x \to +\infty}} = \infty i
     \]

Thus, the limit does not exist in the real number sense and instead involves a complex value.

Would you like more details or explanations on any part? Here are some related questions:

1. How do you handle limits approaching infinity for complex numbers?
2. What happens if you try to calculate the limit in the real number system?
3. How would factoring techniques change if the degree of terms were different?
4. What is the significance of imaginary numbers in calculus limits?
5. How can you simplify similar expressions with complex square roots?

**Tip**: Always check if the expressions inside the square roots lead to complex values when simplifying limits involving large \(x\).

lim as x approaches +∞ of (3x * sqrt(x^2 + x - 3x^2 + x)) / (5 * sqrt(x^2 + 3))

Explore how to find the limit of a complex square root expression as x approaches infinity. This involves simplifying the numerator and denominator and analyzing the behavior of functions at large values of x. Suitable for advanced high school or college-level students.

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