Question

Найти пределы функций

Answer 1

$1)\; \lim\limits _{x \to \infty}\frac{\sqrt[3]{2x^6+1}-x^2+1}{\sqrt{x^4+2}+\sqrt{x^3+1}}=\Big [\frac{:x^2}{:x^2}\Big ]=\lim\limits _{x \to \infty}\frac{\sqrt[3]{2+\frac{1}{x^6}}-1+\frac{1}{x^2}}{\sqrt{1+\frac{2}{x^2}}+\sqrt{\frac{1}{x}+\frac{1}{x^4}}}=\\\\=\frac{\sqrt[3]2-1+0}{1+0}=\sqrt[3]2-1$

$2)\; \lim\limits _{x \to 3}\frac{x^2-5x+6}{x^2-8x+15}=\lim\limits _{x \to 3}\frac{(x-3)(x-2)}{(x-3)(x-5)}=\lim\limits _{x \to \infty}\frac{x-2}{x-5}=\frac{3-2}{3-5}=-0,5\\\\3)\; \lim\limits _{x \to 0}\frac{2\, sinx+arctgx}{tgx-2\, arcsinx}=\lim\limits _{x \to 0}\frac{2\, cosx+\frac{1}{1+x^2}}{\frac{1}{cos^2x}-\frac{2}{\sqrt{1-x^2}}}=\frac{2+1}{1-2}=-3$

$4)\; \star \; \lim\limits _{x \to 0}\Big (\frac{1+x}{2+x}\Big )^{\frac{1-\sqrt{x}}{1-x}}=(\frac{1}{2})^{\frac{1-0}{1+0}}=(\frac{1}{2})^1=\frac{1}{2}\\\\\star \; \star \; \lim\limits _{x \to \infty}\Big (\frac{1+x}{2+x}\Big )^{\frac{1-\sqrt{x}}{1-x}}=[\, 1^{\infty \, }]=\lim\limits _{x \to \infty}\Big ((1+\frac{-1}{x+2})^{\frac{x+2}{-1}}\Big )^{\frac{-1}{x+2}\cdot \frac{1-\sqrt{x}}{(1-\sqrt{x})(1+\sqrt{x})}}=\\\\=e^{\lim\limits _{x \to \infty}\frac{-1}{(x+2)(1+\sqrt{x})}}=\Big [e^{\frac{-1}{\infty }}\Big ]=e^0=1$