Question

Найти предел функций, не пользуясь правилом Лопиталя

Answer 1

$1); limlimits _{x to -infty}(x+sqrt{1+x^2})=limlimits _{x to -infty}frac{x^2-(1+x^2)}{x-sqrt{1+x^2}}=limlimits _{x to -infty}frac{-1}{x-sqrt{1+x^2}}=frac{-1}{-infty }=+0\\2); limlimits _{x to 0}frac{sqrt{1+x}-1}{x}=limlimits _{x to 0}; frac{(1+x)-1}{xcdot (sqrt{1+x}+1)}=limlimits _{x to 0}frac{1}{sqrt{1+x}+1}=frac{1}{2}\\3); limlimits _{x to 1}frac{1-x^2}{sinpi x}=limlimits _{x to 1}frac{1-x^2}{sin(underbrace {pi -pi x}_{to 0})}=Big [, sin, alpha (x)sim alpha (x),; esli; alpha (x)to 0, Big ]=$

$=limlimits _{x to 1}frac{(1-x)(1+x)}{pi -pi x}=limlimits _{x to 1}frac{(1-x)(1+x)}{pi (1-x)}=limlimits _{x to 1}frac{1+x}{pi }=frac{2}{pi }$

$4); limlimits _{x to infty}(frac{2x+1}{2x})^{3x}=limlimits _{x to infty}Big (Big (1+frac{1}{2x}Big )^{2x}Big )^{frac{3x}{2x}}=e^{limlimits _{x to infty}frac{3x}{2x}}=e^{frac{3}{2}}=sqrt{e^3}$