1)lim x-бесконечность -2x-1/3x^2-4x+1 2) lim x-бесконечность 10x^2+4x+3/ 5x^2+2x+1. Created by Malvina247. algebra-ru

5x^2+2x+1...

Malvina247 @Malvina247

August 2021 1 8 Report

1)lim x-бесконечность -2x-1/3x^2-4x+1
2) lim x-бесконечность 10x^2+4x+3/ 5x^2+2x+1

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kirichekov

Verified answer

$\lim_{x \to \infty} \frac{-2x-1}{3 x^{2} -4x+1} = \lim_{x \to \infty} \frac{ \frac{-2x}{ x^{2} }- \frac{1}{ x^{2} } }{ \frac{3 x^{2} }{ x^{2} } - \frac{4x}{ x^{2} } + \frac{1}{ x^{2} } } = \lim_{x \to \infty} \frac{ \frac{-2}{x}- \frac{1}{ x^{2} } }{3- \frac{4}{x} + \frac{1}{ x^{2} } } = \frac{0}{3} =0$
$\frac{-2}{x}=0, \frac{-1}{ x^{2} } =0, \frac{-4}{x} =0$
значение этих величин бесконечно малы при x->∞

$\lim_{x \to \infty} \frac{10 x^{2} +4x+3}{5 x^{2} +2x+1} = \lim_{x \to \infty} \frac{ \frac{10 x^{2} }{ x^{2} }+ \frac{4x}{ x^{2} }+ \frac{3}{ x^{2} } }{ \frac{5 x^{2} }{ x^{2} }+ \frac{2x}{ x^{2} } + \frac{1}{ x^{2} } }= \lim_{x \to \infty} \frac{10+ \frac{4}{x} + \frac{3}{ x^{2} } }{5+ \frac{2}{x} + \frac{1}{ x^{2} } } = \frac{10}{5}=2$

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Answers & Comments

Verified answer

5x^7+X+9...

x^2-9...

pomogite srochno36f8be86f48e9e24971b81cee78cf7a8 46647

pomogite srochno166ded1614878e68dd4886e86ff01464 15644

x-5...

ubytok torgovoj organizacii ot prodazhi tovara na 25 so skidkoj esli roznichnaya

x-9...

5 zadanie srochno8c13aaeed94beca73d649bd691bf6236 93206

koren iz 3f51cb570ff9914bf33f118aa08ce8595 56204

kontent iz 3

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