Question

Вычислить предел, используя формулу тейлора. 25 б

Answer 1

$ln(1+x)=x-\frac{x^2}{2}+o(x^2)\\ \\ \sqrt{1-x}=1-\frac{1}{2}x+o(x)\\ \\x\sqrt{1-x}=x-\frac{1}{2}x^2+o(x^2)\\ \\tgx=x+o(x^2)\\ \\\lim_{x \to 0} \frac{ln(1+x)-x\sqrt{1-x}+\frac{5}{8}x^2}{tgx-x-\frac{x^2}{3}}=\lim_{x \to 0}\frac{x-\frac{x^2}{2} -x+\frac{1}{2}x^2 +\frac{5}{8}x^2+o(x^2)}{x-x-\frac{x^2}{3}+o(x^2)}=\lim_{x \to 0} \frac{\frac{5}{8}x^2+o(x^2)}{-\frac{x^2}{3}+o(x^2}=\frac{\frac{5}{8} }{-\frac{1}{3}}=-\frac{15}{8}$

$cosx=1-\frac{x^2}{2}+o(x^3)\\ \\e^{x}=1+x+\frac{x^2}{2!}+o(x^2)\\ \\ln(1+x)=x-\frac{x^2}{2}+o(x^2)\\ \\\sqrt[3]{8+x}=2\cdot\sqrt[3]{1+\frac{x}{8}}=2\cdot (1+\frac{1}{3}\cdot\frac{x}{8} +\frac{\frac{1}{3} \cdot(\frac{1}{3}-1)}{2!} \frac{x^2}{64}+o(x^2))\\\\ \sqrt[3]{8+x^2}=2\sqrt[3]{1+\frac{x^2}{8}}=2\cdot(1+\frac{1}{3}\cdot\frac{x^2}{8}+o(x^2))$

$\\ \\ \lim_{x \to 0} \frac{cosx-e^{x}+ln(1+x)}{\sqrt[3]{8+x}-\sqrt[3]{8+x^2}+ln12}=\lim_{x \to 0}\frac{1-\frac{x^2}{2}+o(x^3)-1-x-\frac{x^2}{2!}-o(x^2)+x-\frac{x^2}{2}+o(x^2)}{2\cdot (1+\frac{1}{3}\cdot\frac{x}{8} +\frac{\frac{1}{3} \cdot(\frac{1}{3}-1)}{2!} \frac{x^2}{64}+o(x^2))-2\cdot(1+\frac{1}{3}\cdot\frac{x^2}{8}+o(x^2))+ln12}=\\ \\=\lim_{x \to 0}\frac{-3\frac{x^2}{2}+o(x^2)}{-\frac{x^2}{576}+ln12+o(x^2)}=\lim_{x \to 0}\frac{-3\cdot(-576)}{2}=864$