第5章 5.2

(1)

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

(2)

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

(3)

$\lim_{x \to -\infty} \frac{3x+2}{\sqrt{x^2+1}} = \lim_{x \to \infty} \frac{-3x+2}{\sqrt{x^2+1}} = \lim_{x \to \infty} \frac{-3+\dfrac{2}{x}}{\sqrt{1+\dfrac{1}{x^2}}} = -3$

(4)

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

(5)

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

$= \lim_{x \to \infty} \frac{\dfrac{2}{\sqrt{x}}}{\sqrt{1+\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}} = 0$

(6)

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

$= \lim_{x \to \infty} \frac{-1}{x+\sqrt{x^2-1}} = \lim_{x \to \infty} \frac{-\dfrac{1}{x}}{1+\sqrt{1-\dfrac{1}{x^2}}} = 0$