1.3 関数の極限

1.

(a) $${\lim_{x \to 0}(2x-1) = \lim_{x \to 0}2x - \lim_{x \to 0}1 = 0-1 = -1}$$

(b) $${\lim_{x \to 1}\frac{x}{x+1} = \frac{\lim_{x \to 1}x}{\lim_{x \to 1}(x+1)} =\frac{1}{2}}$$

(c) $f(x)$、$g(x)$が多項式で、 $${\lim_{x \to a}\frac{f(x)}{g(x)} = 0}$$となったら、$f(x)$と$g(x)$は因子$(x - a)$をもつ. $${\lim_{x \to 0}\frac{x^{2}(x+1)}{2x} = \lim_{x \to 0}\frac{x(x+1)}{2} = \frac{0}{2} = 0}$$

(d) $${\lim_{x \to 0}\frac{x(x+1)}{2x} = \lim_{x \to 0}\frac{x+1}{2} = \frac{1}{2}}$$

(e) $${\lim_{x \to 9}\frac{x-9}{\sqrt{x} - 3} = \lim_{x \to 9}\frac{(\sqrt{x} + 3)(\sqrt{x} - 3)}{\sqrt{x} - 3} = \lim_{x \to 9}(\sqrt{x} + 3) = 6}$$

(f) $${\lim_{x \to 3}\frac{x^{2} - x - 6}{x - 3} = \lim_{x \to 3}\frac{(x-3)(x+2)}{x - 3} = \lim_{x \to 3}(x+2) = 5}$$

(g) $${\lim_{x \to 4}\frac{(x^{2} - 3x - 4)^{2}}{x - 4} = \lim_{x \to 4}\frac{(x-4)^{2}(x+1)^{2}}{x - 4} = \lim_{x \to 4}(x-4)(x+1)^{2} = 0}$$

(h) $${\lim_{t \to 0}\frac{1}{t}\left(\frac{1}{t+3} - \frac{1}{3}\right)... ...ight) = \lim_{t \to 0}\frac{1}{t}\left(\frac{-t}{3(t+3)}\right) = -\frac{1}{9}}$$

(i) $${\lim_{x \to 0}x\left(1 - \frac{x+1}{x}\right) = \lim_{x \to 0}x\left(\frac{x-(x+1)}{x}\right) = -1}$$

2.

(a) $${\lim_{x \to 0}\frac{\sin{x}}{x} = 1, \lim_{X \to 0}\frac{\sin{X}}{X} = 1}$$より， $${\lim_{x \to 0}\frac{\sin{2x}}{\sin{3x}} = \lim_{x \to 0}\frac{\frac{\sin{2x}}{2x}}{\frac{\sin{3x}}{3x}}\cdot \frac{2x}{3x} = \frac{2}{3}}$$

(b) $${\lim_{x \to 0}\frac{\cos^{2}{x} - 1}{x^{2}} = \lim_{x \to 0}\frac... ...^{2}{x}}{x^{2}} = -\lim_{x \to 0}\frac{\sin{x}}{x}\cdot \frac{\sin{x}}{x} = -1}$$

(c)

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

$$= -\lim_{x \to 0}\frac{\sin{x}}{x} \cdot \frac{\sin{x}}{\cos{x} + 1} = -1 \cdot 0 = 0$$

(d) $x - \pi$を$t$とおくと，$x \to \pi$は$t \to 0$と同値より，

$${\lim_{x \to \pi}\frac{\sin{\pi}}{x - \pi} = \lim_{t \to 0}\frac{\sin{\pi + t)}}{t} = \lim_{t \to 0}\frac{-\sin{t}}{t} = -1}$$