[tex] \lim_{x \to \infty} ( \sqrt{1+x^2} - x)[/tex]. Created by doomla. matematika-ru

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November 2021 1 9 Report

[tex] \lim_{x \to \infty} ( \sqrt{1+x^2} - x)[/tex]

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gartenzie $\lim_{ x \to + \infty }{ ( \sqrt{ 1 + x^2 } - x ) } = \lim_{ x \to + \infty }{ \frac{ ( \sqrt{ 1 + x^2 } - x ) ( \sqrt{ 1 + x^2 } + x ) }{ \sqrt{ 1 + x^2 } + x } } =$

$= \lim_{ x \to + \infty }{ \frac{ ( \sqrt{ 1 + x^2 } )^2 - x^2 }{ \sqrt{ x^2( \frac{1}{x^2} + 1 ) } + x } } = \lim_{ x \to + \infty }{ \frac{ 1 + x^2 - x^2 }{ x \sqrt{ \frac{1}{x^2} + 1 } + x } } =$

$= \lim_{ x \to + \infty }{ \frac{1}{ x ( 1 + \sqrt{ \frac{1}{x^2} + 1 } ) } } = \lim_{ x \to + \infty }{ \frac{1}{x} } \cdot \lim_{ x \to + \infty }{ \frac{1}{ 1 + \sqrt{ \frac{1}{x^2} + 1 } } } =$

$= 0 \cdot \frac{1}{ 1 + \sqrt{ 0 + 1 } } = 0 \cdot \frac{1}{ 1 + \sqrt{1} } = 0 \cdot \frac{1}{ 1 + 1 } = 0 \cdot \frac{1}{2} = 0 \cdot \frac{1}{2} = 0$ ;

$\lim_{ x \to -\infty }{ ( \sqrt{ 1 + x^2 } - x ) } = \lim_{ |x| \to +\infty }{ [ \sqrt{ 1 + |x|^2 } - (-|x|) ] } =$

$= \lim_{ |x| \to +\infty }{ ( \sqrt{ |x|^2( \frac{1}{|x|^2} + 1 ) } + |x| ) } = \lim_{ |x| \to +\infty }{ ( |x| \sqrt{ 1 + \frac{1}{|x|^2} } + |x| ) } =$

$= \lim_{ |x| \to +\infty }{ ( |x| [ 1 + \sqrt{ 1 + \frac{1}{|x|^2} } ] ) } = \lim_{ |x| \to +\infty }{ ( |x| [ 1 + \sqrt{ 1 + 0 } ] ) } =$

$= \lim_{ |x| \to +\infty }{ ( |x| [ 1 + \sqrt{ 1 } ] ) } = \lim_{ |x| \to +\infty }{ ( |x| [ 1 + 1 ] ) } =$

$= \lim_{ |x| \to +\infty }{ ( 2 |x| ) } = +\infty$ ;

О т в е т :

$\lim_{ x \to \infty }{ ( \sqrt{ 1 + x^2 } - x ) } = \left\{\begin{array}{rcl} x \to -\infty & \Rightarrow & = +\infty ; \\x \to +\infty & \Rightarrow & = 0 . \end{array}$

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