Найти указанные пределы, не пользуясь правилом Лопиталя Во вложении.

[latex]1)\; \; \lim\limits _{x\to \infty } \frac{5x^4-6x^2+1}{3x^3-4x^2} =[\, \frac{\infty }{\infty }\, ]=\lim\limits _{x\to \infty } \frac{5-\frac{6}{x^2}+\frac{1}{x^4}}{\frac{3}{x}-\frac{4}{x^2}} =[\, \frac{5-0+0}{0-0}=\frac{5}{0} \, ]=\infty [/latex] [latex]2)\; \; \lim\limits _{x\to 1} \frac{3x^2-2x-1x}{4x^2-3x-1} =[\, \frac{0}{0}\, ]=\lim\limits _{x\to 1} \frac{3(x-1)(x+\frac{1}{3})}{4(x-1)(x+\frac{1}{4})} =\\\\=\frac{3}{4}\lim\limits _{x\to 1 }\frac{x+\frac{1}{3}}{x+\frac{1}{4}}=\frac{3}{4}\cdot \frac{\frac{4}{3}}{\frac{5}{4}}=\frac{4}{5}[/latex] [latex]3)\; \; \lim\limits _{x\to 3} \frac{\sqrt{x+1}-\sqrt{7-x}}{3-x} =[\, \frac{0}{0}\, ]=\lim\limits _{x\to 3} \frac{(\sqrt{x+1}-\sqrt{7-x})(\sqrt{x+1}+\sqrt{7-x})}{(3-x)(\sqrt{x+1}+\sqrt{7-x})} =\\\\=\lim\limits _{x\to 3} \frac{x+1-(7-x)}{(3-x)(\sqrt{x+1}+\sqrt{7-x})} =\lim\limits _{x\to 3} \frac{2(x-3)}{-(x-3)(\sqrt{x+1}+\sqrt{7-x})} =\\\\=-2\cdot \lim\limits _{x\to 3} \frac{1}{\sqrt{x+1}+\sqrt{7-x}} =-2\cdot \frac{1}{2+2}=-\frac{1}{2}[/latex]