### Calculus: Evaluating Limits#### Problem Statement:Solve the limit:\[\lim_{{h \to 0}} \frac{{2\sqrt{{16 + h}} - 8}}{h}\]#### Solution Explanation:To solve the limit \(\lim_{{h \to 0}} \frac{{2\sqrt{{16 + h}} - 8}}{h}\), follow these steps:1. **Substitution Method**: Direct substitution of \( h = 0 \) in the given function \(\frac{2\sqrt{16+h} - 8}{h}\) results in an indeterminate form \(\frac{0}{0}\). Therefore, we need to find another method to simplify the expression.2. **Rationalizing the Numerator**: To eliminate the square root in the numerator, multiply both the numerator and the denominator by the conjugate of the numerator. The conjugate in this case is \(2\sqrt{16+h} + 8\). \[ \frac{2\sqrt{16+h} - 8}{h} \cdot \frac{2\sqrt{16+h} + 8}{2\sqrt{16+h} + 8} \]3. **Simplification**: The product of conjugates will give us a difference of squares in the numerator: \[ = \frac{(2\sqrt{16+h} - 8)(2\sqrt{16+h} + 8)}{h(2\sqrt{16+h} + 8)} \] \[ = \frac{(2\sqrt{16+h})^2 - 8^2}{h(2\sqrt{16+h} + 8)} \] \[ = \frac{4(16+h) - 64}{h(2\sqrt{16+h} + 8)} \] Simplifying inside the numerator: \[ = \frac{64 + 4h - 64}{h(2\sqrt{16+h} + 8)} \] \[ = \frac{4h}{h(2\sqrt{16+h} + 8)} \] Cancelling \( h \) from numerator and denominator: \[ = \frac{4}{

Name: ### Calculus: Evaluating Limits#### Problem Statement:Solve the limit
Uploaded: 2024-03-20T06:38:58.000Z
Channel: Bartleby
Description: ### Calculus: Evaluating Limits#### Problem Statement:Solve the limit:\[\lim_{{h \to 0}} \frac{{2\sqrt{{16 + h}} - 8}}{h}\]#### Solution Explanation:To solve the limit \(\lim_{{h \to 0}} \frac{{2\sqrt{{16 + h}} - 8}}{h}\), follow these steps:1. **Su