1. **State the problem:** We need to evaluate the limit $$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ for each given function and value of $x$. This limit represents the derivative of $f$ at $x$, or the instantaneous rate of change. 2. **Formula and explanation:** The limit definition of the derivative is $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ This gives the slope of the tangent line to the curve at $x$. 3. **Evaluate each problem:** **37.** $f(x) = x^2$, $x=1$ $$ f'(1) = \lim_{h \to 0} \frac{(1+h)^2 - 1^2}{h} = \lim_{h \to 0} \frac{1 + 2h + h^2 - 1}{h} = \lim_{h \to 0} \frac{2h + h^2}{h} = \lim_{h \to 0} (2 + h) = 2 $$ **38.** $f(x) = x^2$, $x=-2$ $$ f'(-2) = \lim_{h \to 0} \frac{(-2+h)^2 - (-2)^2}{h} = \lim_{h \to 0} \frac{4 - 4h + h^2 - 4}{h} = \lim_{h \to 0} \frac{-4h + h^2}{h} = \lim_{h \to 0} (-4 + h) = -4 $$ **39.** $f(x) = 3x - 4$, $x=2$ $$ f'(2) = \lim_{h \to 0} \frac{3(2+h) - 4 - (3 \cdot 2 - 4)}{h} = \lim_{h \to 0} \frac{6 + 3h - 4 - 2}{h} = \lim_{h \to 0} \frac{3h}{h} = 3 $$ **40.** $f(x) = \frac{1}{x}$, $x=-2$ $$ f'(-2) = \lim_{h \to 0} \frac{\frac{1}{-2+h} - \frac{1}{-2}}{h} = \lim_{h \to 0} \frac{\frac{-2 - (-2+h)}{(-2+h)(-2)}}{h} = \lim_{h \to 0} \frac{\frac{-2 + 2 - h}{(-2+h)(-2)}}{h} = \lim_{h \to 0} \frac{-h}{h(-2+h)(-2)} $$ Simplify: $$ = \lim_{h \to 0} \frac{-1}{(-2+h)(-2)} = \frac{-1}{(-2)(-2)} = -\frac{1}{4} $$ **41.** $f(x) = \sqrt{x}$, $x=7$ $$ f'(7) = \lim_{h \to 0} \frac{\sqrt{7+h} - \sqrt{7}}{h} $$ Multiply numerator and denominator by the conjugate: $$ = \lim_{h \to 0} \frac{(\sqrt{7+h} - \sqrt{7})(\sqrt{7+h} + \sqrt{7})}{h(\sqrt{7+h} + \sqrt{7})} = \lim_{h \to 0} \frac{7+h - 7}{h(\sqrt{7+h} + \sqrt{7})} = \lim_{h \to 0} \frac{h}{h(\sqrt{7+h} + \sqrt{7})} = \lim_{h \to 0} \frac{1}{\sqrt{7+h} + \sqrt{7}} $$ Evaluate limit: $$ = \frac{1}{2\sqrt{7}} $$ **42.** $f(x) = \sqrt{3x + 1}$, $x=0$ $$ f'(0) = \lim_{h \to 0} \frac{\sqrt{3(0+h) + 1} - \sqrt{3 \cdot 0 + 1}}{h} = \lim_{h \to 0} \frac{\sqrt{3h + 1} - 1}{h} $$ Multiply numerator and denominator by conjugate: $$ = \lim_{h \to 0} \frac{(\sqrt{3h + 1} - 1)(\sqrt{3h + 1} + 1)}{h(\sqrt{3h + 1} + 1)} = \lim_{h \to 0} \frac{3h + 1 - 1}{h(\sqrt{3h + 1} + 1)} = \lim_{h \to 0} \frac{3h}{h(\sqrt{3h + 1} + 1)} = \lim_{h \to 0} \frac{3}{\sqrt{3h + 1} + 1} $$ Evaluate limit: $$ = \frac{3}{1 + 1} = \frac{3}{2} $$ 4. **Summary of answers:** - 37: $2$ - 38: $-4$ - 39: $3$ - 40: $-\frac{1}{4}$ - 41: $\frac{1}{2\sqrt{7}}$ - 42: $\frac{3}{2}$ These represent the instantaneous rates of change of the functions at the specified points.