1.8 Answer

1.8

1.

(a) $${\lim_{x \to 0} \frac{\log{(1 + x)}}{x} = \lim_{x \to 0}\log{(1+x)^{\frac{1}{x}}} = \log{e} = 1}$$

(b)Let $e^{h} - 1 = x$ . Then $h = \log{(1+x)}$. Thus, $x \to 0$ and $h \to 0$ are equivalent. Therefore, by exercise1-8-1a, we have

$$\lim_{h \to 0} \frac{e^{h} - 1}{h} = \lim_{x \to 0}\frac{x}{\log{(1+x)}} = 1$$

(c)Let $x -a = h$. Then

$$\lim_{x \to a} \frac{\log{x} - \log{a}}{x - a} = \lim_{h \to 0}\frac{\log{(\frac{a+h}{a}}}{h} = \lim_{h \to 0}\frac{\log{(1 + \frac{h}{a}}}{h}$$

$$= \lim_{h \to 0}\log{(1 + \frac{h}{a})^{\frac{1}{h}}} = \frac{1}{a}$$

(d)

$$\lim_{x \to 0}\frac{(1+x)^{\alpha} - 1}{x} = \lim_{x \to 0} \frac{e^{\alpha \log(1+x)} - 1}{x}$$

$$= \lim_{x \to 0}\frac{e^{\alpha \log(1+x)} - 1}{\alpha \log(1+x)}\cdot \frac{\alpha \log(1+x)}{x}$$

$$= \lim_{h \to 0}\frac{e^{h} - 1}{h} \lim_{x \to 0}\frac{\log(1+x)}{x} \alpha = \alpha$$

(e) $${\lim_{x \to 0}x^{x} = \lim_{x \to 0}e^{x \log{x}}}$$. So, we find $\lim_{x \to 0} x \log{x}$．

$$\lim_{x \to 0} x \log{x} = \lim_{x \to 0} \frac{\log{x}}{\frac{1}{x}}$$

Then let $\log{x} = t$ which implies $x = e^{t}$. Thus, $${\lim_{x \to 0} \frac{\log{x}}{\frac{1}{x}} = \lim_{t \to - \infty} \frac{t}{e^{-t}} = 0}$$