Ответ:
[tex]\displaystyle 4)\ 0[/tex]
[tex]\displaystyle 9)\ 0[/tex]
Объяснение:
[tex]\displaystyle 4)\\\\ \lim_{x\to\infty}\frac{2-x^2}{x^3+2x+3}=\lim_{x\to\infty}\frac{\frac{2}{x^3}-\frac{x^2}{x^3}}{\frac{x^3}{x^3}+\frac{2x}{x^3}+\frac{3}{x^3}}=\lim_{x\to\infty}\frac{\frac{2}{x^3}-\frac{1}{x}}{1+\frac{2}{x^2}+\frac{3}{x^3}}=\frac{0}{1}=0[/tex]
[tex]\displaystyle 9)\\\\ \lim_{x\to 2}\frac{x^2-4x+4}{8x^3+4x^2-4x-74}=\frac{2^2-4\cdot 2+4}{8\cdot 2^3+4\cdot 2^2-4\cdot 2-74}=\\\\\\\frac{4-8+4}{8\cdot 8+4\cdot 4-8-74}=0[/tex]
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Answers & Comments
Ответ:
[tex]\displaystyle 4)\ 0[/tex]
[tex]\displaystyle 9)\ 0[/tex]
Объяснение:
[tex]\displaystyle 4)\\\\ \lim_{x\to\infty}\frac{2-x^2}{x^3+2x+3}=\lim_{x\to\infty}\frac{\frac{2}{x^3}-\frac{x^2}{x^3}}{\frac{x^3}{x^3}+\frac{2x}{x^3}+\frac{3}{x^3}}=\lim_{x\to\infty}\frac{\frac{2}{x^3}-\frac{1}{x}}{1+\frac{2}{x^2}+\frac{3}{x^3}}=\frac{0}{1}=0[/tex]
[tex]\displaystyle 9)\\\\ \lim_{x\to 2}\frac{x^2-4x+4}{8x^3+4x^2-4x-74}=\frac{2^2-4\cdot 2+4}{8\cdot 2^3+4\cdot 2^2-4\cdot 2-74}=\\\\\\\frac{4-8+4}{8\cdot 8+4\cdot 4-8-74}=0[/tex]