Behavior near a vertical asymptote

Factor both polynomials: $2x^2 - 5x - 3 = (2x+1)(x-3)$ and $x^2 - 9 = (x-3)(x+3)$. Canceling the common factor $(x-3)$ (valid since $x \ne 3$) gives $f(x) = \dfrac{2x+1}{x+3}$ for $x \neq 3$. This simplified form reveals that near $x=3$, the numerator $2x+1$ approaches $7$ while the denominator $x+3$ approaches $6$, both nonzero.

Since $\dfrac{2x+1}{x+3} \to \dfrac{7}{6}$ as $x \to 3$, and this is a finite nonzero value, the original limit is actually finite, not infinite. Evaluating directly: $\lim_{x\to 3} f(x) = \dfrac{2(3)+1}{3+3} = \dfrac{7}{6}$. Because the factor causing the zero denominator canceled, $x=3$ is a removable discontinuity (a hole), not a vertical asymptote, so the limit exists and equals $\dfrac{7}{6}$.