2.1 Answer

2.1

(a)

$\displaystyle \frac{\cos(3(x+h)) - \cos(3x)}{h} = \frac{\cos{3x}(\cos{3h} - 1)}{h} - \frac{\sin{3x}\sin{3h}}{h} \rightarrow -3\sin{3x}$

(b)

$\displaystyle \frac{(x+2+h)^{n} - (x+2)^n}{h} = \frac{n(x+2)^{n-1}h}{h} + h(\cdots) \rightarrow n(x+2)^{n-1}$

(a)Since $df = f^{\prime}(x) dx$ , we have $df = 4x^3 dx$

(b) $df = e^{x} dx$

(a) $\displaystyle{f_{-}^{'}(0) = \lim_{h \to 0-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0-} \frac{\vert h^2 + h\vert}{h} = \lim_{h \to 0-} \frac{-(h^2 + h)}{h} = -1}$

$\displaystyle{f_{+}^{'}(0) = \lim_{h \to 0+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0+} \frac{\vert h^2 + h\vert}{h} = \lim_{h \to 0+} \frac{h^2 + h}{h} = 1}$

(b) $\displaystyle{f_{-}^{'}(0) = \lim_{h \to 0-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0-} \frac{h^2 \sin{\frac{1}{h}}}{h} = \lim_{h \to 0-} h \sin{\frac{1}{h}} = 0}$

$\displaystyle{f_{+}^{'}(0) = \lim_{h \to 0+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0+} \frac{h^2 \sin{\frac{1}{h}}}{h} = \lim_{h \to 0+} h \sin{\frac{1}{h}} = 0}$

(c) $\displaystyle{f_{-}^{'}(0) = \lim_{h \to 0-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0-} \frac{\sqrt{h^3 + h^2}}{h} = \lim_{h \to 0-} \sqrt{1 + h} = 1}$

$\displaystyle{f_{+}^{'}(0) = \lim_{h \to 0+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0+} \frac{\sqrt{h^3 + h^2}}{h} = \lim_{h \to 0+} \sqrt{1 + h} = 1}$