3) \( \lim _{x \rightarrow 6} \frac{x^{2}+3 x-54}{x^{2}-36}= \)

To solve the limit \( \lim _{x \rightarrow 6} \frac{x^{2}+3 x-54}{x^{2}-36} \), we can start by substituting \( x = 6 \) directly into the function: 1. The numerator becomes \( 6^2 + 3(6) - 54 = 36 + 18 - 54 = 0 \). 2. The denominator becomes \( 6^2 - 36 = 36 - 36 = 0 \). Since both the numerator and the denominator approach zero, we have a \( \frac{0}{0} \) indeterminate form. Therefore, we need to factor both the numerator and the denominator. **Factoring the numerator**: \[ x^2 + 3x - 54 = (x - 6)(x + 9) \] **Factoring the denominator**: \[ x^2 - 36 = (x - 6)(x + 6) \] Now we can rewrite the limit: \[ \lim_{x \rightarrow 6} \frac{(x-6)(x+9)}{(x-6)(x+6)} \] We can cancel the common factor \( (x - 6) \) from the numerator and denominator (keeping in mind that this cancellation is valid as long as \( x \neq 6 \)): \[ = \lim_{x \rightarrow 6} \frac{x + 9}{x + 6} \] Now, we can substitute \( x = 6 \): \[ = \frac{6 + 9}{6 + 6} = \frac{15}{12} = \frac{5}{4} \] Thus, the limit is: \[ \lim _{x \rightarrow 6} \frac{x^{2}+3 x-54}{x^{2}-36} = \frac{5}{4} \]