Question

Помогите решить предел( в задании требуют использовать правило Лопиталя)

Answer 1

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

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