1. **Problem:** Find the limit \(\lim_{x \to 5} 8x - 2\). 2. **Formula and rule:** For limits of polynomial or linear functions, the limit as \(x\) approaches a value is simply the function evaluated at that value. 3. **Calculate values for the table:** - \(f(4.9) = 8(4.9) - 2 = 39.2 - 2 = 37.2\) - \(f(4.91) = 8(4.91) - 2 = 39.28 - 2 = 37.28\) - \(f(4.911) = 8(4.911) - 2 = 39.288 - 2 = 37.288\) - \(f(5) = 8(5) - 2 = 40 - 2 = 38\) - \(f(5.001) = 8(5.001) - 2 = 40.008 - 2 = 38.008\) - \(f(5.002) = 8(5.002) - 2 = 40.016 - 2 = 38.016\) - \(f(5.003) = 8(5.003) - 2 = 40.024 - 2 = 38.024\) 4. **Conclusion:** The limit is \(\lim_{x \to 5} 8x - 2 = 38\). 5. **Problem:** Find the limit \(\lim_{x \to 2} \frac{9 + x}{2x}\). 6. **Formula and rule:** For rational functions where the denominator is not zero at the point, the limit is the function evaluated at that point. 7. **Calculate values for the table:** - \(f(1.97) = \frac{9 + 1.97}{2(1.97)} = \frac{10.97}{3.94} \approx 2.784\) - \(f(1.98) = \frac{9 + 1.98}{2(1.98)} = \frac{10.98}{3.96} \approx 2.773\) - \(f(1.998) = \frac{9 + 1.998}{2(1.998)} = \frac{10.998}{3.996} \approx 2.75\) - \(f(2) = \frac{9 + 2}{2(2)} = \frac{11}{4} = 2.75\) - \(f(2.003) = \frac{9 + 2.003}{2(2.003)} = \frac{11.003}{4.006} \approx 2.746\) - \(f(2.02) = \frac{9 + 2.02}{2(2.02)} = \frac{11.02}{4.04} \approx 2.727\) - \(f(2.1) = \frac{9 + 2.1}{2(2.1)} = \frac{11.1}{4.2} \approx 2.643\) 8. **Conclusion:** The limit is \(\lim_{x \to 2} \frac{9 + x}{2x} = 2.75\).