Question

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

Answer 1

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

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

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

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


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

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

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

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

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



О т в е т :

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