Question

Вычислите предел по правилу Лопиталя

Answer 1

$limlimits _{x to infty}frac{e^{frac{1}{x^2}}-1}{2arctg^2x-pi }=limlimits _{x to infty}frac{frac{1}{x^2}}{2cdot (frac{pi}{2})^2-pi }= limlimits _{x to infty}frac{frac{1}{x^2}}{frac{pi ^2}{2}-pi }=Big [; frac{0}{const}; Big ]=0\\\(e^{alpha (x)}-1)sim a(x); ; ,; ; esli; alpha (x)to 0; ; ,; alpha (x)=frac{1}{x^2}to 0; pri; xto infty$

$limlimits _{x to infty}frac{e^{frac{1}{x^2}}-1}{2arctg(x^2)-pi }=[; frac{1-1}{2cdot frac{pi}{2}-pi }=frac{0}{0}; ,; Lopital]=limlimits _{x to infty}frac{e^{frac{1}{x^2}}cdot (-2x^{-3})}{2cdot frac{2x}{1+x^4}}=\\=limlimits _{x to infty}frac{-2cdot e^{frac{1}{x^2}}}{x^3cdot frac{4x}{1+x^4}}=-limlimits _{x to infty}frac{e^{frac{1}{x^2}}cdot (1+x^4)}{2x^4} =[; Lopital; ]=$

$=Big [; e^{frac{1}{x^2}}to e^0=1; ,; (1+x^4)to infty ; ,; ; (1+x^4)sim x^4; ,; (2x^4)to infty ; Big ]=\\=-limlimits _{x to infty}frac{x^4}{2x^4}=-frac{1}{2}$