Ако някой може да помогне

[Гост]

1642060157884.jpg (1.76 MiB) Прегледано 305 пъти

Моля Ви за помощ за тези две задачи
Благодаря

[ammornil]

[tex]\lim_{x \to -1}\frac{x^{3}-2x-1}{x^{5}-2x-1}=\lim_{x \to -1}\frac{x^{3}-x-x-1}{x^{5}-x-x-1}=\lim_{x \to -1}\frac{x(x^{2}-1)-(x+1)}{x(x^{4}-1)-(x+1)}=\lim_{x \to -1}\frac{x(x-1)(x+1)-(x+1)}{x(x^{2}-1)(x^{2}+1)-(x+1)}=[/tex]

[tex]=\lim_{x \to -1}\frac{(x+1)[x(x-1)-1]}{x(x-1)(x+1)(x^{2}+1)-(x+1)}=\lim_{x \to -1}\frac{\cancel{(x+1)}[x(x-1)-1]}{\cancel{(x+1)}[x(x-1)(x^{2}+1)-1]}=...=\frac{(-1)(-1-1)}{(-1).(-1-1)[(-1)^{2}+1]-1}=[/tex]

[tex]=\frac{-1.(-2)}{-1.(-2).2-1}=\frac{2}{3}[/tex]

[ammornil]

ще изполваме, че [tex]\lim_{x \to \infty }\frac{a^{x}}{b^{x}}=0[/tex] когато [tex]a<b[/tex].

[tex]\lim_{x \to \infty }\frac{3^{x}+7^{x}}{7^{x+2}-5^{x}} = \lim_{x \to \infty }\frac{7^{x}(\frac{3^{x}}{7^{x}}+1)}{7^{2}.7^{x}-5^{x}}= \lim_{x \to \infty }\frac{\cancel{7^{x}}(\frac{3^{x}}{7^{x}}+1)}{\cancel{7^{x}}(7^{2}-\frac{5^{x}}{7^{x}})}=...=\frac{0+1}{49-0}=\frac{1}{49}[/tex]

[Гост]

Благодаря!!!!