1. The problem requires finding $\delta$ for each limit statement given $f(x)$, $x_0$, $L = \lim_{x \to x_0} f(x)$, and $\varepsilon$. We want $\delta > 0$ s.t. if $0 < |x - x_0| < \delta$ then $|f(x) - L| < \varepsilon$.\\ 2. For exercise 31: $f(x) = 3 - 2x$, $x_0 = 3$, $\varepsilon = 0.02$. - Limit: $L = 3 - 2(3) = 3 - 6 = -3$. - Solve $|f(x) - L| < \varepsilon$: $$|3 - 2x + 3| < 0.02 \implies |-2x + 6| < 0.02$$ $$2|x - 3| < 0.02 \implies |x - 3| < \frac{0.02}{2} = 0.01$$ So $\delta = 0.01$. 3. Exercise 32: $f(x) = -3x - 2$, $x_0 = -1$, $\varepsilon = 0.03$. - Limit: $L = -3(-1) - 2 = 3 - 2 = 1$ - $|-3x - 2 - 1| < 0.03 \implies |-3x - 3| < 0.03$ - $3|x + 1| < 0.03 \implies |x + 1| < 0.01$ - $\delta = 0.01$. 4. Exercise 33: $f(x) = \frac{x^2 - 4}{x-2}$ at $x_0 = 2$, $\varepsilon = 0.05$. - Simplify: $f(x) = \frac{(x-2)(x+2)}{x-2} = x + 2$, $x \ne 2$. - Limit: $L = 2 + 2 = 4$. - $|f(x) - 4| < 0.05 \implies |x + 2 - 4| < 0.05 \implies |x - 2| < 0.05$. - So $\delta = 0.05$. 5. Exercise 34: $f(x) = \frac{x^2 + 6x + 5}{x + 5}$, $x_0 = -5$, $\varepsilon = 0.05$. - Factor numerator: $(x + 1)(x + 5)$. - Simplify for $x \ne -5$: $f(x) = x + 1$. - Limit: $L = -5 + 1 = -4$. - $|f(x) - L| < \varepsilon \implies |x + 1 + 4| < 0.05 \implies |x + 5| < 0.05$. - $\delta = 0.05$. 6. Exercise 35: $f(x) = \sqrt{1 - 5x}$, $x_0 = -3$, $\varepsilon = 0.5$. - Limit: $L = \sqrt{1 - 5(-3)} = \sqrt{1 + 15} = \sqrt{16} = 4$. - Solve $|\sqrt{1-5x} - 4| < 0.5$. - Square inequality bounds: - Lower bound: $4 - 0.5 =3.5$, Upper bound: $4 + 0.5 = 4.5$. - So $3.5 < \sqrt{1-5x} < 4.5$. - Square both: - $3.5^2 < 1 - 5x < 4.5^2$. - $12.25 < 1 - 5x < 20.25$. - Solve inequalities: - Left: $1 - 5x > 12.25 \implies -5x > 11.25 \implies x < -\frac{11.25}{5} = -2.25$. - Right: $1 - 5x < 20.25 \implies -5x < 19.25 \implies x > -\frac{19.25}{5} = -3.85$. - So $-3.85 < x < -2.25$. - Since $x_0 = -3$, distance to endpoints: min$(-3 + 3.85, -2.25 + 3) = \min(0.85, 0.75) = 0.75$. - $\delta = 0.75$. 7. Exercise 36: $f(x) = \frac{4}{x}$, $x_0 = 2$, $\varepsilon = 0.4$. - Limit: $L = \frac{4}{2} = 2$. - Solve $|\frac{4}{x} - 2| < 0.4$. - $|\frac{4 - 2x}{x}| < 0.4$. - To avoid complexity, impose $x$ near 2, e.g. $|x - 2| < 1$ so $x \in (1, 3)$. - For $x \in (1,3)$, $|x| \ge 1$. - Multiply inequality by $|x| \ge 1$: $|4 - 2x| < 0.4|x|$ and since $|x| < 3$, $|4 - 2x| < 1.2$. - $|4 - 2x| = 2|2 - x| < 1.2 \implies |x - 2| < 0.6$. - So $\delta = 0.6$. Final answers summarized: - 31: $\delta=0.01$ - 32: $\delta=0.01$ - 33: $\delta=0.05$ - 34: $\delta=0.05$ - 35: $\delta=0.75$ - 36: $\delta=0.6$. This solves the first 6 exercises on formal limits with given $\varepsilon$.