!!!!!!!!!!!!Предел функции! за решение 30 пунктов!!!!!!!!!!!!!!. Created by YALO001. algebra-ru

!!!!!!!!!!!!Предел функции! за решение 30 пунктов!!!!!!!!!!!!!!...

Nik133

$1)\ \lim_{x \to \infty} \frac{3+x+5x^4}{x^4-12x+1}=[\frac{\infty}{\infty}]= \lim_{x \to \infty} \frac{5x^4}{x^4}=5 \\ \\ \\ 2)\ \lim_{x \to 0} \frac{\sqrt{1+3x}-\sqrt{1-3x}}{x+x^2}= \\ \\ = \lim_{x \to 0} \frac{(\sqrt{1+3x}-\sqrt{1-3x})(\sqrt{1+3x}+\sqrt{1-3x})}{(x+x^2)(\sqrt{1+3x}+\sqrt{1-3x})}= \\ \\ =\lim_{x \to 0} \frac{1+3x-1+3x}{x(1+x)(\sqrt{1+3x}+\sqrt{1-3x})}= \\ \\ =\lim_{x \to 0} \frac{6}{(1+x)(\sqrt{1+3x}+\sqrt{1-3x})}=\frac{6}{(1+0)(1+1)}=\frac{6}{2}=3$

$3)\ \lim_{x \to \pi/4} \frac{cos^2x}{sin^3x}=\frac{cos^2(\pi/4)}{sin^3(\pi/4)}=(\frac{\sqrt2}{2})^2 :(\frac{\sqrt2}{2})^3=\frac{2}{4}*\frac{8}{2\sqrt2}= \\ \\ =\frac{2}{\sqrt2}=\frac{2\sqrt2}{\sqrt2*\sqrt2}=\frac{2\sqrt2}{2}=\sqrt2$

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