To determine the limit as $x \to \infty$ for the given rational function:

$\lim_{x \to \infty} \frac{x^2 - \frac{3}{5}x^3 + 7x - 5}{x^2 - \frac{5}{3}x + \frac{6}{5}x^3}$

Step 1: Analyze the highest degree terms

When dealing with limits of rational functions, the behavior as $x \to \infty$ is dominated by the highest degree terms in the numerator and denominator.

• Numerator: $x^2 - \frac{3}{5}x^3 + 7x - 5$.
  The highest degree term is $-\frac{3}{5}x^3$.

• Denominator: $x^2 - \frac{5}{3}x + \frac{6}{5}x^3$.
  The highest degree term is $\frac{6}{5}x^3$.

Step 2: Factor out the highest power of $x$ (i.e., $x^3$)

Factor $x^3$ from both numerator and denominator:

$\frac{x^2 - \frac{3}{5}x^3 + 7x - 5}{x^2 - \frac{5}{3}x + \frac{6}{5}x^3} = \frac{\left(-\frac{3}{5}x^3 + \text{lower degree terms}\right)}{\left(\frac{6}{5}x^3 + \text{lower degree terms}\right)}.$

Dividing through by $x^3$:

$\frac{-\frac{3}{5} + \frac{x^2}{x^3} + \frac{7x}{x^3} - \frac{5}{x^3}}{\frac{6}{5} + \frac{x^2}{x^3} - \frac{5}{3x^3} + \frac{6}{5x^3}}.$

As $x \to \infty$, the lower degree terms (those divided by $x^3$) approach zero.

Step 3: Simplify the leading terms

The limit simplifies to:

$\frac{-\frac{3}{5}}{\frac{6}{5}}.$

Simplify the fraction:

$\frac{-\frac{3}{5}}{\frac{6}{5}} = -\frac{3}{5} \cdot \frac{5}{6} = -\frac{3}{6} = -\frac{1}{2}.$

Step 4: Verify the options

From the given choices:

• A: $-3/2$
• B: $-15/4$
• C: $0$
• D: $1$
• E: $5/2$

None of the options seem to match the correct answer. Let me know if you'd like me to double-check or further clarify any steps.