\( A(4+h, a) \) on \( f(x)=\sqrt{x}+1 \) \( B(4, b) \) \( \lim _{h \rightarrow 0} \frac{2 a-2 b}{h}= \) ?

To solve this limit, we first need to find the values of \( a \) and \( b \) by substituting \( x = 4 \) into the function \( f(x) = \sqrt{x} + 1 \). Calculating \( b \): \[ b = f(4) = \sqrt{4} + 1 = 2 + 1 = 3. \] Next, for point \( A(4+h, a) \), we compute: \[ a = f(4+h) = \sqrt{4+h} + 1. \] Now, we can rewrite the expression for the limit as: \[ \lim_{h \to 0} \frac{2a - 2b}{h} = \lim_{h \to 0} \frac{2(\sqrt{4+h} + 1) - 2(3)}{h} = \lim_{h \to 0} \frac{2\sqrt{4+h} + 2 - 6}{h} = \lim_{h \to 0} \frac{2\sqrt{4+h} - 4}{h}. \] Factoring out \( 2 \): \[ \lim_{h \to 0} \frac{2(\sqrt{4+h} - 2)}{h}. \] Applying L'Hôpital's Rule or simplifying the expression: As \( h \to 0, \sqrt{4+h} \to 2 \), we find: \[ \frac{2(\sqrt{4+h} - 2)}{h} \text{ becomes } \frac{2(0)}{0} \text{ so we can apply L'Hôpital's Rule.} \] Differentiating the numerator and denominator: Numerator: \( \frac{d}{dh}(2(\sqrt{4+h} - 2)) = 2 \cdot \frac{1}{2\sqrt{4+h}} = \frac{1}{\sqrt{4+h}}. \) Denominator: \( \frac{d}{dh}(h) = 1. \) Thus: \[ \lim_{h \to 0} \frac{\frac{1}{\sqrt{4+h}}}{1} = \frac{1}{\sqrt{4}} = \frac{1}{2}. \] So the result is: \[ \lim _{h \rightarrow 0} \frac{2 a - 2 b}{h} = 1. \]