Question

Помогите решить предел( в задании требуют использовать правило Лопиталя)

Answer 1

$displaystyle lim_{x to infty}left(xln x-sqrt{1+x^2}right)=lim_{x to infty}dfrac{x^2ln^2x-1-x^2}{xln x+sqrt{1+x^2}}=lim_{x to infty}dfrac{(x^2ln^2x-1-x^2)'_x}{(xln x+sqrt{1+x^2})'_x}\ \ \ =lim_{x to infty}dfrac{2xln^2x+x^2cdot frac{2ln x}{x}-2x}{ln x+xcdot frac{1}{x}+frac{2x}{2sqrt{1+x^2}}}=lim_{x to infty}frac{(2xln^2x+2xln x-2x)'_x}{(ln x+1+frac{x}{sqrt{1+x^2}})'_x}=$

$=displaystyle lim_{x to infty}dfrac{2ln^2x+2xcdotfrac{2ln x}{x}-2}{frac{1}{x}+frac{sqrt{1+x^2}-xcdot frac{x}{sqrt{1+x^2}}}{1+x^2}}=2lim_{x to infty}dfrac{ln^2x+2ln x-1}{frac{1}{x}+frac{1+x^2-x^2}{(1+x^2)sqrt{1+x^2}}}=\ \ \ =2lim_{x to infty}frac{ln^2x+2ln x-1}{frac{1}{x}+frac{1}{(1+x^2)sqrt{1+x^2}}}=2cdot dfrac{lim_{x to infty}(ln^2 x+2ln x-1)}{0}=infty$