Question

Найти пределы функций

Answer 1

$1)\; \; \lim\limits _{x \to 3}\frac{2x^2-5x-3}{3x^2-4x-15}=\lim\limits _{x \to 3}\frac{2(x+\frac{1}{2})(x-3)}{3(x+\frac{5}{3})(x-3)}=\lim\limits _{x \to 3}\frac{2x+1}{3x+5}=\frac{2\cdot 3+1}{3\cdot 3+5}=\frac{7}{14} =0,5\\\\\\\lim\limits _{x \to \infty }\frac{2x^2-5x-3}{3x^2-4x-15}=\lim\limits _{x \to \infty}\frac{2-\frac{5}{x}-\frac{3}{x^2}}{3-\frac{4}{x}-\frac{15}{x^2}}=\frac{2}{3}\\\\\\2)\; \; \lim\limits _{x \to 0}\frac{\sqrt{x+4}-2}{x}=\lim\limits _{x \to 0}\frac{(x+4)-4}{x\, (\sqrt{x+4}+2)}=\lim\limits _{x \to 0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2+2}=0,25$

$3)\; \; \lim\limits _{x \to 0}\frac{1-cos^24x}{4x}=\lim\limits _{x \to 0}\frac{sin^24x}{4x}=[\; sin\alpha \sim \alpha \; ,\; esli\; \; \alpha \to 0\; ]=\\\\=\lim\limits_{x \to 0}\frac{(4x)^2}{4x}=\lim\limits_{x \to 0}(4x)=0\\\\\\4)\; \; \lim\limits _{x \to \infty}\Big (\frac{x+9}{x-3}\Big )^{x}=\Big [\, 1^{\infty }\, \Big ]=\lim\limits_{x \to \infty}\Big (\Big (1+\frac{12}{x-3}\Big )^{\frac{x-3}{12}}\Big )^{\frac{12\, x}{x-3}}=\\\\\\=e^{\lim\limits _{x \to \infty}\frac{12\, x}{x-3}}=e^{12}$