1.4 Answer

1.4

1.

(a) $${\lim_{x \to 0-} \frac{1}{x} = - \infty}$$ and $${\lim_{x \to 0+} \frac{1}{x} = \infty}$$． Thus, the limit does not exist.

(b) $${\lim_{x \to 0} \frac{1}{x^2} = \infty}$$

(c)

$$\lim_{x \to 0+} \frac{\vert x\vert}{\sqrt{a+x} - \sqrt{a - x}} = \lim_{x \to 0+} \frac{x(\sqrt{a +x} + \sqrt{a-x})}{\sqrt{a+x} - \sqrt{a - x})(\sqrt{a+x} + \sqrt{a-x})}$$

$$= \lim_{x \to 0+}\frac{x(\sqrt{a+x} + \sqrt{a-x})}{a+x - (a-x)} = \sqrt{a}$$

(d)

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

$$= \lim_{x \to 0-}\frac{x \sqrt{1 + \cos{x}}}{\vert\sin{x}\vert} = \lim_{x \to 0-}\frac{x \sqrt{1 + \cos{x}}}{-\sin{x}} = - \sqrt{2}$$

(e) $${\lim_{x \to \infty} \frac{1}{x} = 0}$$ ，Since $\cos{x}$ is continuous， $${\lim_{x \to \infty} \cos{\frac{1}{x}} = \cos(\lim_{x \to \infty} \frac{1}{x}) = 1}$$

(f) $${\lim_{x \to \infty} \sin{x}}$$ vibrates. Since $0 \leq \vert\sin{x}\vert \leq 1$. Thus， $${0 \leq \vert\frac{\sin{x}}{x}\vert \leq \frac{1}{x}}$$.

Therefore, $${\lim_{x \to \infty} \frac{\sin{x}}{x} = 0}$$

(g) $${\lim_{x \to 0+} \frac{1}{x} = \infty}$$ suggest that $${\lim_{x \to 0+} \cos(\frac{1}{x})}$$ does not exist. To show this, let $t = \frac{1}{x}$. Then $x \to 0+$ implies $t \to \infty$. Then $${\lim_{x \to 0+} \cos(\frac{1}{x}) = \lim_{t \to \infty} \cos(t)}$$and the limit does not exist.