Question

Помогите пожалуйста решить 2 вариант . 2 ,3,4,5

Answer 1

$lim_{x to 0} frac{x}{ sqrt{3+x} - sqrt{3-x} } = lim_{x to 0} frac{x( sqrt{3+x} + sqrt{3-x} )}{ (sqrt{3+x})^2 - (sqrt{3-x})^2 } =\ =lim_{x to 0} frac{x( sqrt{3+x} + sqrt{3-x} )}{ 3+x-3+x } =lim_{x to 0} frac{x( sqrt{3+x} + sqrt{3-x} )}{2x }=\ = lim_{x to 0} frac{ sqrt{3+x} + sqrt{3-x} }{2 }= frac{ sqrt{3}+ sqrt{3} }{2} = sqrt{3}$

$lim_{x to 0} frac{1- sqrt{1-x^2} }{x^2}= lim_{x to 0} frac{1^2- (sqrt{1-x^2})^2 }{x^2(1+ sqrt{1-x^2})}= lim_{x to 0} frac{1-1+x^2 }{x^2(1+ sqrt{1-x^2})}=\ = lim_{x to 0} frac{x^2 }{x^2(1+ sqrt{1-x^2})}=lim_{x to 0} frac{1 }{1+ sqrt{1-x^2}}= frac{1}{1+ sqrt{1} } =0,5$

$lim_{x to infty} frac{5x^4-x^3+2x}{x^4-8x^3+1}= lim_{x to infty} frac{x^4( 5-frac{1}{x}+ frac{2}{x^3})}{x^4(1- frac{8}{x}+ frac{1}{x^4} )}=5$

$lim_{x to infty} (1+ frac{3}{x} )^{-x}= lim_{x to infty} ((1+ frac{3}{x} )^{ frac{x}{3} })^{-x* frac{3}{x} }=\=lim_{x to infty} ((1+ frac{3}{x} )^{ frac{x}{3} })^{-3 }=e^{-3}$