无题

$$v_正 = k_1 \cdot c(I_2)$$ $$v_逆 = k_2 \cdot c^2(I)$$ $$\because v_正 = v_逆$$ $$\therefore k_1 \cdot c(I_2) = k_2 \cdot c^2(I)$$ $$\therefore c^2(I) = \frac{k_1 \cdot c(I_2)}{k_2}$$ $$\therefore v = k_3 \cdot c(H_2) \cdot c^2(I) = k_3 \cdot c(H_2) \cdot \frac{k_1 \cdot c(I_2)}{k_2} = \frac{k_3 \cdot k_1}{k_2} \cdot c(H_2) \cdot c(I_2)$$ $$v = k \cdot c(H_2) \cdot c(I_2) \qquad k = \frac{k_3 \cdot k_1}{k_2}$$ $$\displaystyle \lim_{x \to 0}{\frac{(1+x)^\frac{2}{x}-e^2[1-ln(1+x)]}{x}}$$ $$\displaystyle \lim_{x \to \infty}{sin^2(\pi \sqrt{n^2+n})}$$ $$\displaystyle \lim_{x \to 0}{\frac{(1-\frac{1}{2}x^2)^\frac{2}{3}-1}{xln(1+x)}}$$