1. **State the problem:** Find the limits $ \lim_{x \to 9} \frac{3 + \sqrt{x}}{x - 9} $ and $ \lim_{t \to -1} \frac{t^2 - 3t - 4}{t^3 - 1} $. 2. **First limit:** $ \lim_{x \to 9} \frac{3 + \sqrt{x}}{x - 9} $. - Direct substitution gives denominator zero: $9 - 9 = 0$, so we need to simplify. - Use substitution $ \sqrt{x} = y \Rightarrow x = y^2 $, then as $ x \to 9 $, $ y \to 3 $. - Rewrite denominator: $ x - 9 = y^2 - 9 = (y - 3)(y + 3) $. - The expression becomes $ \frac{3 + y}{(y - 3)(y + 3)} $. - Notice numerator $3 + y$ and denominator factor $y + 3$ are the same, so cancel: $$ \frac{3 + y}{(y - 3)(y + 3)} = \frac{\cancel{3 + y}}{(y - 3)\cancel{(y + 3)}} = \frac{1}{y - 3} $$ - Now find $ \lim_{y \to 3} \frac{1}{y - 3} $, which tends to infinity (does not exist as finite). 3. **Second limit:** $ \lim_{t \to -1} \frac{t^2 - 3t - 4}{t^3 - 1} $. - Factor numerator: $ t^2 - 3t - 4 = (t - 4)(t + 1) $. - Factor denominator: $ t^3 - 1 = (t - 1)(t^2 + t + 1) $. - Substitute $ t = -1 $ directly: Numerator: $ (-1 - 4)(-1 + 1) = (-5)(0) = 0 $ Denominator: $ (-1 - 1)((-1)^2 + (-1) + 1) = (-2)(1 - 1 + 1) = (-2)(1) = -2 $ - So the limit is $ \frac{0}{-2} = 0 $. **Final answers:** - $ \lim_{x \to 9} \frac{3 + \sqrt{x}}{x - 9} $ does not exist (tends to infinity). - $ \lim_{t \to -1} \frac{t^2 - 3t - 4}{t^3 - 1} = 0 $.