Question

Salut, aveti o idee la problema numarul 325 ??

Answer 1

$\displaystyle x- \sqrt{x^2+x+1} \cdot \frac{\ln(e^x+x)}{x}= \\ \\ = x- \sqrt{x^2+x+1} \cdot \frac{x+ \ln \left(1+ \frac{x}{e^x} \right)}{x}= \\ \\ = x- \sqrt{x^2+x+1} \cdot \left(1+ \frac{\ln \left(1+ \frac{x}{e^x} \right)}{x} \right)=\\ \\ =x- \sqrt{x^2+x+1}+ \frac{\sqrt{x^2+x+1}}{x} \cdot \ln \left(1+ \frac{x}{e^x} \right). \\ \\ -----------$

$\displaystyle x- \sqrt{x^2+x+1}= \frac{x^2-(x^2+x+1)}{x+ \sqrt{x^2+x+1}}=- \frac{x+1}{x+ \sqrt{x^2+x+1}}= \\ \\=- \frac{1+\frac{1}{x}}{1+ \sqrt{1+ \frac{1}{x}+ \frac{1}{x^2}}} \to- \frac{1}{2}. \\ \\ \frac{\sqrt{x^2+x+1}}{x} \cdot \ln \left(1+ \frac{x}{e^x} \right) = \sqrt{1+\frac{1}{x}+ \frac{1}{x^2}} \ln \left(1+ \frac{x}{e^x} \right) \to \\ \\ \to  1 \cdot \ln 1 =0. \\ \\ Prin~urmare~limita~este~- \frac{1}{2}.$