4)) Evaluate the limit by rationalizing the denominator \( \lim _{t \rightarrow-5} \frac{2 t+10}{\sqrt{18 t+171}-\sqrt{16 t+161}} \) 1)) Simplify any fractions and radicals in your answer. Submit

To evaluate the limit \[ \lim _{t \rightarrow-5} \frac{2 t+10}{\sqrt{18 t+171}-\sqrt{16 t+161}}, \] we start by substituting \( t = -5 \). 1. Calculating expressions: \[ 2(-5) + 10 = 0, \] \[ \sqrt{18(-5) + 171} = \sqrt{-90 + 171} = \sqrt{81} = 9, \] \[ \sqrt{16(-5) + 161} = \sqrt{-80 + 161} = \sqrt{81} = 9. \] So we see that as \( t \to -5 \), the denominator approaches \( 9 - 9 = 0 \) and the numerator approaches \( 0 \). Thus, we have the indeterminate form \( \frac{0}{0} \). 2. We rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, \( \sqrt{18t + 171} + \sqrt{16t + 161} \): \[ \frac{(2t + 10)(\sqrt{18t + 171} + \sqrt{16t + 161})}{(\sqrt{18t + 171} - \sqrt{16t + 161})(\sqrt{18t + 171} + \sqrt{16t + 161})}. \] The denominator simplifies to: \[ (\sqrt{18t + 171})^2 - (\sqrt{16t + 161})^2 = (18t + 171) - (16t + 161) = 2t + 10. \] 3. Now the expression becomes: \[ \frac{(2t + 10)(\sqrt{18t + 171} + \sqrt{16t + 161})}{2t + 10}. \] Since \( t \neq -5 \) in the limit, we can cancel \( 2t + 10 \): \[ \sqrt{18t + 171} + \sqrt{16t + 161}. \] 4. Next, we can substitute \( t = -5 \): \[ \sqrt{18(-5) + 171} + \sqrt{16(-5) + 161} = \sqrt{81} + \sqrt{81} = 9 + 9 = 18. \] 5. Thus, the limit is: \[ \lim_{t \to -5} \frac{2 t+10}{\sqrt{18 t+171} - \sqrt{16 t+161}} = 18. \] So the final answer is: \[ \boxed{18}. \]