Question

Cрочно, только выделенные

Answer 1

2.
$lim_{x to infty} ( frac{2}{x^2}+ frac{3}{x}-1)* frac{7x^2-4}{2x^2+3}= \ \ = lim_{x to infty} ( frac{2}{x^2}+ frac{3}{x}-1 )* lim_{x to infty} frac{7x^2-4}{2x^2+3}= \ \ =-1* lim_{x to infty} frac{ frac{7x^2}{x^2}- frac{4}{x^2} }{ frac{2x^2}{x^2}+ frac{3}{x^2} }=-1* lim_{x to infty} frac{7- frac{4}{x^2} }{2+ frac{3}{x^2} }= \ \ =-1* frac{7}{2}=-3.5$

5.
a)
$g=( frac{2}{x}-1 )cosx \ \ g'=( frac{2}{x}-1 )'cosx+( frac{2}{x}-1 )(cosx)'= \ \ =( -frac{2}{x^2} )cosx-( frac{2}{x}-1 )sinx= \ \ =- frac{2cosx}{x^2}- frac{2sinx}{x}+sinx= frac{x^2sinx-2cosx-2xsinx}{x^2}$

6.
$y= frac{( sqrt{x} -3)sinx}{x^2}=x^{-2}( sqrt{x} -3)sinx= \ \ =(x^{-2} sqrt{x} -3x^{-2})sinx=(x^{-2+0.5}-3x^{-2})sinx= \ \ =(x^{-1.5}-3x^{-2})sinx$

$y'=(x^{-1.5}-3x^{-2})'sinx+(x^{-1.5}-3x^{-2})(sinx)'= \ \ =(-1.5x^{-2.5}+6x^{-3})sinx-(x^{-1.5}-3x^{-2})cosx= \ \ =(- frac{3}{2x^2 sqrt{x} }+ frac{6}{x^3} )sinx-( frac{1}{x sqrt{x} }- frac{3}{x^2} )cosx= \ \ =( frac{6}{x^3}- frac{3 sqrt{x} }{2x^3} )sinx-( frac{ sqrt{x} -3}{x^2} )cosx= \ \ =( frac{12-3 sqrt{x} }{2x^3} )sinx-( frac{ sqrt{x} -3}{x^2} )cosx$

Answer 2

Спасибо большое)