câu hỏi đúng sai

Câu 2: a) Ta có: \[ f(x) = 3x + 1 \] Khi \( x \to +\infty \), ta thấy rằng \( 3x \to +\infty \). Do đó: \[ \lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (3x + 1) = +\infty \] b) Ta có: \[ \lim_{x \to -\infty} \frac{f(x)}{3 - 5x} = \lim_{x \to -\infty} \frac{3x + 1}{3 - 5x} \] Chia cả tử và mẫu cho \( x \): \[ \lim_{x \to -\infty} \frac{3x + 1}{3 - 5x} = \lim_{x \to -\infty} \frac{3 + \frac{1}{x}}{\frac{3}{x} - 5} \] Khi \( x \to -\infty \), ta có \( \frac{1}{x} \to 0 \) và \( \frac{3}{x} \to 0 \). Do đó: \[ \lim_{x \to -\infty} \frac{3 + \frac{1}{x}}{\frac{3}{x} - 5} = \frac{3 + 0}{0 - 5} = \frac{3}{-5} = -\frac{3}{5} \] c) Ta có: \[ \lim_{x \to (-\frac{1}{3})^+} \frac{f(x)}{3x^2 - 2x - 1} = \lim_{x \to (-\frac{1}{3})^+} \frac{3x + 1}{3x^2 - 2x - 1} \] Chia cả tử và mẫu cho \( x + \frac{1}{3} \): \[ \lim_{x \to (-\frac{1}{3})^+} \frac{3x + 1}{3x^2 - 2x - 1} = \lim_{x \to (-\frac{1}{3})^+} \frac{3(x + \frac{1}{3})}{(x + \frac{1}{3})(3x - 3)} \] \[ = \lim_{x \to (-\frac{1}{3})^+} \frac{3}{3x - 3} \] Khi \( x \to (-\frac{1}{3})^+ \), ta có \( 3x - 3 \to -3 - 3 = -6 \). Do đó: \[ \lim_{x \to (-\frac{1}{3})^+} \frac{3}{3x - 3} = \frac{3}{-6} = -\frac{1}{2} \] d) Ta có: \[ \lim_{x \to 0} \frac{\sqrt[3]{f(x)} - \sqrt{2x + 1}}{3x^2} = \lim_{x \to 0} \frac{\sqrt[3]{3x + 1} - \sqrt{2x + 1}}{3x^2} \] Áp dụng công thức nhân liên hợp: \[ \lim_{x \to 0} \frac{\sqrt[3]{3x + 1} - \sqrt{2x + 1}}{3x^2} = \lim_{x \to 0} \frac{(\sqrt[3]{3x + 1} - \sqrt{2x + 1})(\sqrt[3]{(3x + 1)^2} + \sqrt[3]{3x + 1}\sqrt{2x + 1} + \sqrt{(2x + 1)^2})}{3x^2 (\sqrt[3]{(3x + 1)^2} + \sqrt[3]{3x + 1}\sqrt{2x + 1} + \sqrt{(2x + 1)^2})} \] \[ = \lim_{x \to 0} \frac{(3x + 1) - (2x + 1)}{3x^2 (\sqrt[3]{(3x + 1)^2} + \sqrt[3]{3x + 1}\sqrt{2x + 1} + \sqrt{(2x + 1)^2})} \] \[ = \lim_{x \to 0} \frac{x}{3x^2 (\sqrt[3]{(3x + 1)^2} + \sqrt[3]{3x + 1}\sqrt{2x + 1} + \sqrt{(2x + 1)^2})} \] \[ = \lim_{x \to 0} \frac{1}{3x (\sqrt[3]{(3x + 1)^2} + \sqrt[3]{3x + 1}\sqrt{2x + 1} + \sqrt{(2x + 1)^2})} \] Khi \( x \to 0 \), ta có \( \sqrt[3]{(3x + 1)^2} \to 1 \), \( \sqrt[3]{3x + 1}\sqrt{2x + 1} \to 1 \), và \( \sqrt{(2x + 1)^2} \to 1 \). Do đó: \[ \lim_{x \to 0} \frac{1}{3x (1 + 1 + 1)} = \lim_{x \to 0} \frac{1}{9x} = \frac{1}{6} \] Đáp số: a) \( +\infty \) b) \( -\frac{3}{5} \) c) \( -\frac{1}{2} \) d) \( \frac{1}{6} \)