1.3 Answer

1.3

1.

(a)

$$\lim_{x \to 2}(x^2 + 4x) = \lim_{x \to 2}x^2 + \lim_{x \to 2}4x = 4 + 8 = 12$$

(b) As $x \to 2$，both $x^2 - 3x + 2$ and $x^2 - 4$ approach to 0. Thus,

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

(c)

$$\lim_{x \rightarrow 0} \frac{2x^4 - 6x^3 + x^2 + 2}{x - 1} = \frac{2}{-1} = -2$$

(d)

$$\lim_{x \rightarrow 1} \frac{2x^4 - 6x^3 + x^2 + 3}{x - 1} = \lim_{x \rightarrow 1}\frac{(x-1)(2x^3 - 4x^2 - 3x - 3)}{x - 1} = -8$$

(e)

$$\lim_{x \rightarrow 0} \frac{\sqrt{1+x} - \sqrt{1 - x}}{x} = \lim_{x \rightarrow 0} \frac{(\sqrt{1+x} - \sqrt{1 - x})(\sqrt{1+x} + \sqrt{1 - x})}{x(\sqrt{1+x} + \sqrt{1 - x})}$$

$$= \lim_{x \rightarrow 0} \frac{{1+x} - (1 - x)}{x(\sqrt{1+x} + \sqrt{1 - x})}$$

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

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

(f) We use $a - b = \frac{a^3 - b^3}{a^2 + ab + b^2}$．

$$\lim_{x \rightarrow 0}\frac{(1+x)^{1/3} - (1 - x)^{1/3}}{x} = \lim_{x \rightarrow 0}\frac{(1+x) - (1 - x)}{x[(1+x)^{2/3} + (1+x)^{1/3}(1-x)^{1/3} + (1-x)^{2/3}]}$$

$$= \lim_{x \rightarrow 0}\frac{2x}{x[(1+x)^{2/3} + (1+x)^{1/3}(1-x)^{1/3} + (1-x)^{2/3}]}$$

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

(g)

$$\lim_{x \to 2}f(x) = \lim_{x \to 2}x^2 = 4$$

2.

(a)

$$\lim_{x \to 0} \frac{\sin{2x}}{\sin{3x}} = \lim_{x \to 0} \frac{2... ...dot \frac{\lim_{x \to 0} \sin{2x}/2x}{\lim_{x \to 0} \sin{3x}/3x} = \frac{2}{3}$$

(b)

$$\lim_{x \rightarrow 0}\frac{\cos{x} - 1}{x} = \lim_{x \rightarrow 0}\frac{(\cos{x} - 1)(\cos{x} + 1)}{x(\cos{x} + 1)}$$

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

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

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

Alternate solution $${\frac{1 - \cos{x}}{2} = \sin^{2}{\frac{x}{2}}}$$を用いると

$$\lim_{x \rightarrow 0}\frac{\cos{x} - 1}{x} = \lim_{x \rightarrow 0} \frac{-2 sin^{2}{\frac{x}{2}}}{x}$$

$$= -2\lim_{x \to 0} \frac{\sin^{2}{\frac{x}{2}}}{2 \frac{x}{2}}$$

$$= -4\lim_{x \to 0} \sin{\frac{x}{2}} \cdot \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} = 0$$

(c) Let $y = \sin^{-1}{x}$. Then $x = \sin{y}$．Also，for $x \to 0$，$y \to 0$．Thus,

$$\lim_{x \rightarrow 0}\frac{\sin^{-1}{x}}{x} = \lim_{y \rightarrow 0}\frac{y}{\sin{y}} = 1$$

(d)

$$0 \leq \vert x \sin{\frac{1}{x}}\vert = \vert x\vert\vert\sin{\frac{1}{x}}\vert \leq \vert x\vert$$

Then by squeezing theorem,

$$\lim_{x \to 0} \vert x \sin{\frac{1}{x}}\vert = 0$$

Thus，

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

3.

Note that $\vert\vert f(x)\vert - 0\vert = \vert f(x) - 0\vert$．For all $\varepsilon > 0$，there exists $\delta$ such that for $\vert x - a\vert < \delta$, then $\vert\vert f(x)\vert - 0\vert = \vert f(x) - 0\vert < \varepsilon$