Các mệnh đề sau đúng hay sai?

a) Đúng b) Đúng c) Đúng d) Sai Lời giải a) Đúng. Ta có: $$ \lim_{x \to 0} \frac{\sqrt{4 + 3x} - 2}{4x} = \lim_{x \to 0} \frac{(\sqrt{4 + 3x} - 2)(\sqrt{4 + 3x} + 2)}{4x(\sqrt{4 + 3x} + 2)} $$ $$ = \lim_{x \to 0} \frac{4 + 3x - 4}{4x(\sqrt{4 + 3x} + 2)} = \lim_{x \to 0} \frac{3}{4(\sqrt{4 + 3x} + 2)} = \frac{3}{4(\sqrt{4} + 2)} = \frac{3}{16} $$. b) Đúng. Ta có: $$ \lim_{x \to +\infty} \frac{3x^2 - 4x + 1}{2x^2 + 2x + 1} = \lim_{x \to +\infty} \frac{x^2 \Big(3 - \frac{4}{x} + \frac{1}{x^2}\Big)}{x^2 \Big(2 + \frac{2}{x} + \frac{1}{x^2}\Big)} $$ $$ = \lim_{x \to +\infty} \frac{3 - \frac{4}{x} + \frac{1}{x^2}}{2 + \frac{2}{x} + \frac{1}{x^2}} = \frac{3}{2} $$. c) Đúng. Ta có: $$ \lim_{x \to 3} \frac{\sqrt{2x + 3} - 3}{\sqrt{x + 1} - 2} = \lim_{x \to 3} \frac{(\sqrt{2x + 3} - 3)(\sqrt{2x + 3} + 3)(\sqrt{x + 1} + 2)}{(\sqrt{x + 1} - 2)(\sqrt{x + 1} + 2)(\sqrt{2x + 3} + 3)} $$ $$ = \lim_{x \to 3} \frac{(2x + 3 - 9)(\sqrt{x + 1} + 2)}{(x + 1 - 4)(\sqrt{2x + 3} + 3)} = \lim_{x \to 3} \frac{2(\sqrt{x + 1} + 2)}{\sqrt{2x + 3} + 3} = \frac{4}{3} $$ d) Sai. Ta có: $$ \lim_{x \to 1} \frac{\sqrt[3]{x + 7} - 2}{x^2 - 3x + 2} = \lim_{x \to 1} \frac{(\sqrt[3]{x + 7} - 2)\Big(\sqrt[3]{(x+7)^2} + 2\sqrt[3]{x+7} + 4\Big)}{(x^2 - 3x + 2)\Big(\sqrt[3]{(x+7)^2} + 2\sqrt[3]{x+7} + 4\Big)} $$ $$ = \lim_{x \to 1} \frac{x + 7 - 8}{(x-1)(x-2)\Big(\sqrt[3]{(x+7)^2} + 2\sqrt[3]{x+7} + 4\Big)} = \lim_{x \to 1} \frac{1}{(x-2)\Big(\sqrt[3]{(x+7)^2} + 2\sqrt[3]{x+7} + 4\Big)} = -\frac{1}{12} $$.