1. **State the problem:** Find the limit as $x$ approaches infinity of the expression $$\frac{\sqrt{13x^2 + 5x + 1}}{2x - 3}.$$\n\n2. **Recall the formula and rules:** When dealing with limits involving square roots and polynomials as $x \to \infty$, we focus on the highest degree terms because they dominate the behavior.\n\n3. **Simplify the expression:**\nWe factor out $x^2$ inside the square root:\n$$\sqrt{13x^2 + 5x + 1} = \sqrt{x^2 \left(13 + \frac{5}{x} + \frac{1}{x^2}\right)} = |x| \sqrt{13 + \frac{5}{x} + \frac{1}{x^2}}.$$\nSince $x \to \infty$, $|x| = x$. So,\n$$\sqrt{13x^2 + 5x + 1} = x \sqrt{13 + \frac{5}{x} + \frac{1}{x^2}}.$$\n\n4. **Rewrite the original limit:**\n$$\lim_{x \to \infty} \frac{x \sqrt{13 + \frac{5}{x} + \frac{1}{x^2}}}{2x - 3} = \lim_{x \to \infty} \frac{x \sqrt{13 + \frac{5}{x} + \frac{1}{x^2}}}{x \left(2 - \frac{3}{x}\right)} = \lim_{x \to \infty} \frac{\sqrt{13 + \frac{5}{x} + \frac{1}{x^2}}}{2 - \frac{3}{x}}.$$\n\n5. **Evaluate the limit:** As $x \to \infty$, terms with $\frac{1}{x}$ and $\frac{1}{x^2}$ approach 0, so\n$$\lim_{x \to \infty} \frac{\sqrt{13 + 0 + 0}}{2 - 0} = \frac{\sqrt{13}}{2}.$$\n\n**Final answer:** $$\boxed{\frac{\sqrt{13}}{2}}.$$