Limits Continuity 504Dd0

1. **Problem Statement:** Find the limits and continuity properties as requested in the exercises. --- ### Exercise 1.1 (4) Given the graph description: (a) $\lim_{x\to 2^-} f(x)$: The left branch approaches an open circle at $(2, 2)$, so the left-hand limit is $2$. (b) $\lim_{x\to 2^+} f(x)$: The right branch starts at a filled point at $(2, 2)$ and flattens near $y=2$, so the right-hand limit is $2$. (c) $\lim_{x\to 2} f(x)$: Since left and right limits are equal, the limit exists and equals $2$. (d) $f(2)$: The filled point at $(2, 2)$ means $f(2) = 2$. --- ### Exercise 1.2 (5) Find $\lim_{x\to 3} \frac{x^2 - 2x}{x + 1}$ 1. Substitute $x=3$ directly: $$\frac{3^2 - 2\cdot 3}{3 + 1} = \frac{9 - 6}{4} = \frac{3}{4}$$ So, the limit is $\frac{3}{4}$. (9) Find $\lim_{x\to 1} \frac{x^2 + 6x + 5}{x^2 - 3x - 4}$ 1. Factor numerator and denominator: $$x^2 + 6x + 5 = (x + 1)(x + 5)$$ $$x^2 - 3x - 4 = (x - 4)(x + 1)$$ 2. Simplify the fraction: $$\frac{(x + 1)(x + 5)}{(x - 4)(x + 1)} = \frac{\cancel{(x + 1)}(x + 5)}{(x - 4)\cancel{(x + 1)}} = \frac{x + 5}{x - 4}$$ 3. Substitute $x=1$: $$\frac{1 + 5}{1 - 4} = \frac{6}{-3} = -2$$ So, the limit is $-2$. (29) Find $\lim_{x\to 9} \frac{x - 9}{\sqrt{x} - 3}$ 1. Direct substitution gives $\frac{0}{0}$, so use algebraic manipulation. 2. Multiply numerator and denominator by the conjugate $\sqrt{x} + 3$: $$\frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} = \frac{(x - 9)(\sqrt{x} + 3)}{x - 9}$$ 3. Cancel $x - 9$: $$\frac{\cancel{(x - 9)}(\sqrt{x} + 3)}{\cancel{x - 9}} = \sqrt{x} + 3$$ 4. Substitute $x=9$: $$\sqrt{9} + 3 = 3 + 3 = 6$$ So, the limit is $6$. --- ### Exercise 1.3 (14) Find $\lim_{x\to \infty} \frac{5x^2 - 4x}{2x^2 + 3}$ 1. Divide numerator and denominator by $x^2$: $$\frac{5 - \frac{4}{x}}{2 + \frac{3}{x^2}}$$ 2. As $x \to \infty$, terms with $\frac{1}{x}$ and $\frac{1}{x^2}$ go to $0$: $$\frac{5 - 0}{2 + 0} = \frac{5}{2}$$ So, the limit is $\frac{5}{2}$. (17) Find $\lim_{x\to -\infty} \frac{x - 2}{x^2 + 2x + 1}$ 1. The denominator is $x^2 + 2x + 1 = (x + 1)^2$ which grows like $x^2$. 2. Divide numerator and denominator by $x^2$: $$\frac{\frac{x}{x^2} - \frac{2}{x^2}}{1 + \frac{2}{x} + \frac{1}{x^2}} = \frac{\frac{1}{x} - \frac{2}{x^2}}{1 + \frac{2}{x} + \frac{1}{x^2}}$$ 3. As $x \to -\infty$, $\frac{1}{x} \to 0$, $\frac{2}{x^2} \to 0$: $$\frac{0 - 0}{1 + 0 + 0} = 0$$ So, the limit is $0$. (23) Find $\lim_{x\to +\infty} \sqrt[3]{\frac{2 + 3x - 5x^2}{1 + 8x^2}}$ 1. Divide numerator and denominator inside the cube root by $x^2$: $$\sqrt[3]{\frac{\frac{2}{x^2} + \frac{3}{x} - 5}{\frac{1}{x^2} + 8}}$$ 2. As $x \to +\infty$, terms with $\frac{1}{x}$ and $\frac{1}{x^2}$ go to $0$: $$\sqrt[3]{\frac{0 + 0 - 5}{0 + 8}} = \sqrt[3]{\frac{-5}{8}} = -\sqrt[3]{\frac{5}{8}} = -\frac{\sqrt[3]{5}}{2}$$ So, the limit is $-\frac{\sqrt[3]{5}}{2}$. (25) Find $\lim_{x\to -\infty} \frac{\sqrt{5x^2 - 2}}{x + 3}$ 1. For large $|x|$, $\sqrt{5x^2 - 2} \approx \sqrt{5} |x|$. 2. Since $x \to -\infty$, $|x| = -x$. 3. Rewrite numerator: $$\sqrt{5x^2 - 2} = \sqrt{5} |x| = \sqrt{5} (-x) = -\sqrt{5} x$$ 4. So the expression is approximately: $$\frac{-\sqrt{5} x}{x + 3}$$ 5. Divide numerator and denominator by $x$: $$\frac{-\sqrt{5} \cancel{x}}{\cancel{x} + \frac{3}{x}} = \frac{-\sqrt{5}}{1 + 0} = -\sqrt{5}$$ So, the limit is $-\sqrt{5}$. --- ### Exercise 1.5 (15) Find values of $x$ where $f(x) = \frac{x}{2x + 2}$ is not continuous. 1. The function is undefined where denominator is zero: $$2x + 2 = 0 \implies x = -1$$ 2. So, $f$ is not continuous at $x = -1$. (29a) Find $k$ so that $$f(x) = \begin{cases} 7x - 2, & x \leq 1 \\ kx^2, & x > 1 \end{cases}$$ is continuous everywhere. 1. Continuity at $x=1$ requires: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$ 2. Compute left limit and value: $$f(1) = 7(1) - 2 = 5$$ 3. Compute right limit: $$\lim_{x \to 1^+} kx^2 = k(1)^2 = k$$ 4. Set equal for continuity: $$k = 5$$ So, $k = 5$ makes $f$ continuous everywhere. --- **Final answers:** - Exercise 1.1 (4): (a) 2, (b) 2, (c) 2, (d) 2 - Exercise 1.2 (5): $\frac{3}{4}$ - Exercise 1.2 (9): $-2$ - Exercise 1.2 (29): 6 - Exercise 1.3 (14): $\frac{5}{2}$ - Exercise 1.3 (17): 0 - Exercise 1.3 (23): $-\frac{\sqrt[3]{5}}{2}$ - Exercise 1.3 (25): $-\sqrt{5}$ - Exercise 1.5 (15): Not continuous at $x = -1$ - Exercise 1.5 (29a): $k = 5$