1. **State the problem:** We want to find the limit as $h \to 0$ of the difference quotient for $f(x) = 4x + x^2$ at $x=2$, which is $$\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}.$$ Also, we want to evaluate the limit involving $g(x) = \frac{1}{3x-7}$ as $h \to 0$ at $x=2$. 2. **Recall the difference quotient formula:** $$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a),$$ where $f'(a)$ is the derivative of $f$ at $a$. This limit represents the slope of the tangent line to $f$ at $x=a$. 3. **Calculate $f(2+h)$ and $f(2)$:** $$f(2+h) = 4(2+h) + (2+h)^2 = 8 + 4h + 4 + 4h + h^2 = 12 + 8h + h^2,$$ $$f(2) = 4(2) + 2^2 = 8 + 4 = 12.$$ 4. **Form the difference quotient:** $$\frac{f(2+h) - f(2)}{h} = \frac{12 + 8h + h^2 - 12}{h} = \frac{8h + h^2}{h}.$$ 5. **Simplify by canceling $h$:** $$\frac{\cancel{h}(8 + h)}{\cancel{h}} = 8 + h.$$ 6. **Take the limit as $h \to 0$:** $$\lim_{h \to 0} (8 + h) = 8.$$ 7. **Interpretation:** The derivative $f'(2) = 8$, so the limit is correct for the first function. 8. **Now evaluate the limit for $g(x) = \frac{1}{3x - 7}$ at $x=2$: $$\lim_{h \to 0} \frac{1}{3(2+h) - 7} + 1 = \lim_{h \to 0} \frac{1}{6 + 3h - 7} + 1 = \lim_{h \to 0} \frac{1}{-1 + 3h} + 1.$$ 9. **Evaluate the limit:** As $h \to 0$, denominator approaches $-1$, so $$\lim_{h \to 0} \frac{1}{-1 + 3h} + 1 = \frac{1}{-1} + 1 = -1 + 1 = 0.$$ 10. **What was done wrong?** - In the first limit, you wrote the expansion incorrectly: $(2+h)^2 = 4 + 4h + h^2$, but you added $4h$ twice. The correct expansion is $4 + 4h + h^2$. - In the second limit, you must substitute carefully and simplify before taking the limit. Your final evaluation is correct. **Final answer:** $$\lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = 8,$$ $$\lim_{h \to 0} \frac{1}{3(2+h) - 7} + 1 = 0.$$