Evaluating a limit by factoring to remove a 0/0 form

Since substituting $x=2$ directly gives $\frac{4+6-10}{0} = \frac{0}{0}$, an algebraic simplification is needed. Factor the numerator: $x^2 + 3x - 10 = (x-2)(x+5)$. This shows the numerator and denominator share a common factor of $(x-2)$.

For $x \neq 2$, we can cancel the common factor to write $a(x) = x+5$. Since this simplified expression is a polynomial (continuous everywhere), the quotient and sum rules justify evaluating the limit by direct substitution: $\lim_{x\to 2}(x+5) = 2+5 = 7$.

After factoring the numerator and canceling the common factor of x minus two, the expression simplified to x plus five. This linear expression is continuous everywhere, so direct substitution of x equals two could then be applied. The cancellation step is what removed the original zero over zero indeterminate form.