Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.
f(x) = √(x + 3); P (1,2)

Verified step by step guidance

1. Step 1: Recall the definition of the derivative as the slope of the tangent line at a point. The derivative of a function f at a point x = a is given by the limit: \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \).
2. Step 2: Identify the function and the point of tangency. Here, the function is \( f(x) = \sqrt{x + 3} \) and the point P is (1, 2).
3. Step 3: Substitute the point of tangency into the derivative definition. We need to find \( f'(1) \), so substitute \( a = 1 \) into the limit: \( f'(1) = \lim_{h \to 0} \frac{\sqrt{1+h+3} - \sqrt{1+3}}{h} \).
4. Step 4: Simplify the expression inside the limit. This becomes \( \lim_{h \to 0} \frac{\sqrt{h+4} - 2}{h} \). To simplify further, multiply the numerator and the denominator by the conjugate of the numerator: \( \frac{\sqrt{h+4} - 2}{h} \times \frac{\sqrt{h+4} + 2}{\sqrt{h+4} + 2} \).
5. Step 5: Simplify the resulting expression. The numerator becomes \( (\sqrt{h+4})^2 - 2^2 = h + 4 - 4 = h \), so the expression simplifies to \( \lim_{h \to 0} \frac{h}{h(\sqrt{h+4} + 2)} \). Cancel \( h \) in the numerator and denominator, and evaluate the limit as \( h \to 0 \).