To find the limit as \( x \) approaches 3 for the expression
\[ \frac{x^2 + 2x - 15}{x - 3}, \]
we first substitute \( x = 3 \) directly into the expression:
\[ \frac{3^2 + 2(3) - 15}{3 - 3} = \frac{9 + 6 - 15}{0} = \frac{0}{0}. \]
This indicates that we have an indeterminate form, so we will factor the numerator to simplify the expression.
The numerator \( x^2 + 2x - 15 \) can be factored. We need two numbers that multiply to \(-15\) and add to \(2\), which are \(5\) and \(-3\):
\[ x^2 + 2x - 15 = (x - 3)(x + 5). \]
Now, we can rewrite the limit:
\[ \frac{(x - 3)(x + 5)}{x - 3}. \]
Provided \( x \neq 3 \), we can cancel \( x - 3 \):
\[ x + 5. \]
Now we find the limit as \( x \) approaches \( 3 \):
\[ \lim_{x \to 3} (x + 5) = 3 + 5 = 8. \]
Thus, the limit is
\[ \boxed{8}. \]