The given problem involves finding the limit of the expression as $x \to \infty$:

$\lim_{{x \to \infty}} \frac{\sqrt{3x} - 2}{\sqrt{2 - \frac{3}{5}x}}$

Steps to solve the limit:

1. Analyze the behavior of the numerator:

   • The numerator is $\sqrt{3x} - 2$.
   • As $x \to \infty$, $\sqrt{3x}$ grows without bound, and the constant $-2$ becomes negligible in comparison. Therefore, the numerator approaches $\sqrt{3x} \sim \sqrt{3} \cdot \sqrt{x}$.

2. Analyze the behavior of the denominator:

   • The denominator is $\sqrt{2 - \frac{3}{5}x}$.
   • As $x \to \infty$, the term $-\frac{3}{5}x$ dominates over the constant $2$, so the denominator behaves like $\sqrt{-\frac{3}{5}x} \sim \sqrt{\frac{3}{5}} \cdot \sqrt{x}$.
   • Notice that the expression becomes undefined unless we take into account the correct interpretation: the term inside the square root is negative, so we likely have a typographical issue or an absolute value is implied to keep the expression valid.

Approximation as $x \to \infty$:

• Numerator: $\sqrt{3x} \sim \sqrt{3} \cdot \sqrt{x}$.
• Denominator: $\sqrt{-\frac{3}{5}x} \sim \sqrt{\frac{3}{5}} \cdot \sqrt{x}$, but due to negative square root, the limit might not exist.

If we assume the sign is adjusted (absolute value taken or reconsider the context), this simplifies the problem into: $\frac{\sqrt{3} \cdot \sqrt{x}}{\sqrt{\frac{3}{5}} \cdot \sqrt{x}} = \frac{\sqrt{3}}{\sqrt{\frac{3}{5}}} = \sqrt{5}.$

Thus, the limit is: $\boxed{\sqrt{5}}.$

Do you want further details or have any questions?

Related Questions:

1. What is the meaning of an indeterminate form in limits?
2. How can we handle square roots of negative numbers in real analysis?
3. What alternative methods can be used to solve limits involving radicals?
4. How does L'Hopital's rule apply to limits of this type?
5. What happens if the denominator becomes zero in a limit?

Tip: Always verify if the expression inside square roots stays positive as $x \to \infty$, especially when dealing with negative terms.