第5章 5.18

(1)

$\lim_{x\to\infty} \sqrt{x}(\sqrt{x+1}-\sqrt{x}) = \lim_{x\to\infty} \frac{\sqrt{x}(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}} = \lim_{x\to\infty} \frac{\sqrt{x}(x+1-x)}{\sqrt{x+1}+\sqrt{x}} = \lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}$

$= \lim_{x\to\infty} \frac{1}{\sqrt{1+\frac{1}{x}}+1} = \frac{1}{1+1} = \frac{1}{2}$

(2)

$\lim_{x\to 0} \frac{\sqrt{1+2x+3x^2}-\sqrt{1-3x}}{3x} = \lim_{x\to 0} \frac{(\sqrt{1+2x+3x^2}-\sqrt{1-3x})(\sqrt{1+2x+3x^2}+\sqrt{1-3x})}{3x(\sqrt{1+2x+3x^2}+\sqrt{1-3x})}$

$= \lim_{x\to 0} \frac{1+2x+3x^2-(1-3x)}{3x(\sqrt{1+2x+3x^2}+\sqrt{1-3x})} = \lim_{x\to 0} \frac{5x+3x^2}{3x(\sqrt{1+2x+3x^2}+\sqrt{1-3x})} = \lim_{x\to 0} \frac{5+3x}{3(\sqrt{1+2x+3x^2}+\sqrt{1-3x})}$

$= \frac{5}{3(\sqrt{1}+\sqrt{1})} = \frac{5}{3\cdot 2} = \frac{5}{6}$

(3)

$\lim_{x\to\infty} (\sqrt{x^2+x+1}-\sqrt{x^2+1}) = \lim_{x\to\infty} \frac{(\sqrt{x^2+x+1}-\sqrt{x^2+1})(\sqrt{x^2+x+1}+\sqrt{x^2+1})}{(\sqrt{x^2+x+1}+\sqrt{x^2+1})}$

$= \lim_{x\to\infty} \frac{(x^2+x+1-(x^2+1))}{(\sqrt{x^2+x+1}+\sqrt{x^2+1})} = \lim_{x\to\infty} \frac{x}{(\sqrt{x^2+x+1}+\sqrt{x^2+1})} = \lim_{x\to\infty} \frac{x}{\left(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x^2}}\right)}$

$= \frac{1}{1+1} = \frac{1}{2}$