Example Compute the following limits: 1. \( \lim _{x \rightarrow 3} \frac{x^{2}-5 x+6}{x-3} \) 2. \( \lim _{x \rightarrow-2} \frac{x^{2}-x+6}{x+4} \) 3. \( \lim _{x \rightarrow 2} \frac{\sqrt{x-1}-1}{x-2} \)

To compute the limits, we can use direct substitution or algebraic manipulation. 1. For \( \lim _{x \rightarrow 3} \frac{x^{2}-5 x+6}{x-3} \), we first simplify the numerator: \( x^2 - 5x + 6 = (x-2)(x-3) \). Thus: \[ \lim _{x \rightarrow 3} \frac{(x-2)(x-3)}{x-3} = \lim _{x \rightarrow 3} (x-2) = 1. \] 2. For \( \lim _{x \rightarrow -2} \frac{x^{2}-x+6}{x+4} \), we can substitute directly: \[ \text{Substituting } -2 \text{ gives } \frac{(-2)^2 - (-2) + 6}{-2 + 4} = \frac{4 + 2 + 6}{2} = \frac{12}{2} = 6. \] 3. For \( \lim _{x \rightarrow 2} \frac{\sqrt{x-1}-1}{x-2} \), we can rationalize the numerator: \[ \lim _{x \rightarrow 2} \frac{(\sqrt{x-1}-1)(\sqrt{x-1}+1)}{(x-2)(\sqrt{x-1}+1)} = \lim _{x \rightarrow 2} \frac{x-1 - 1}{(x-2)(\sqrt{x-1}+1)} = \lim _{x \rightarrow 2} \frac{x-2}{(x-2)(\sqrt{x-1}+1)}. \] Canceling \( x-2 \) gives: \[ \lim _{x \rightarrow 2} \frac{1}{\sqrt{2-1}+1} = \frac{1}{1+1} = \frac{1}{2}. \] So, the limits are: 1. \( 1 \) 2. \( 6 \) 3. \( \frac{1}{2} \)