Намерете границите на функциите:

[Sweetbubble]

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

[tex]\lim_{x \to 3} \frac{x^{2}-9}{\sqrt{x}-\sqrt{4x-9}}[/tex]

[Nathi123]

[tex]A= \lim_{x \to -\infty}\frac{-5x^{4}+2x^{2}+1}{3x+2} = \lim_{x \to -\infty}\frac{x^{4}(-5+\frac{2}{x^{2}}+\frac{1}{x^{4}})}{x(3 + \frac{2}{x})}[/tex]
[tex]\lim_{x \to -\infty}\frac{2}{x^{2}}=\lim_{x \to -\infty}\frac{1}{x^{4}}=\lim_{x \to -\infty}\frac{2}{x}=0\Rightarrow A= \lim_{x \to -\infty}x^{3}.\frac{-5}{3} = +\infty[/tex]
B=[tex]\lim_{x \to 3} \frac{x^{2}-9}{\sqrt{x} - \sqrt{4x-9}} = \lim_{x \to 3}\frac{(x-3)(x+3)(\sqrt{x}+\sqrt{4x-9})}{(\sqrt{x}-\sqrt{4x-9})(\sqrt{x}+\sqrt{4x-9}}=\lim_{x \to 3}\frac{(x-3)(x+3)(\sqrt{x}+\sqrt{4x-9})}{x-4x+9}[/tex]
=[tex]\frac{1}{3}\lim_{x \to 3}\frac{(x - 3)(x + 3).2\sqrt{3}}{(3-x)}= -\frac{2\sqrt{3}}{3}.\lim_{x \to 3}(x+3)=-4\sqrt{3}.[/tex]