Question

Salut, ma puteti ajuta la problema numarul 1230, nu il gasesc pe acel F(x), am incercat sa derivez f(x) insa nu iasa...

Answer 1

$\displaystyle \boxed{1230}~\lim_{x \to 0} \frac{F(x)}{\frac{f(x)}{x}}=0,~deoarece: \\ \\ \lim_{x \to 0}F(x)=F(0); \\ \\ \lim_{x \to 0} \frac{f(x)}{x}=\lim_{x \to 0} \frac{e^{x^2}}{x}= +\infty. \\ \\ \boxed{1231} \lim_{x \to \infty} \frac{F(x)}{\frac{f(x)}{x}} \\ \\ Se~stie~ca~e^x \ge x+1~\forall~x \ge 0. \\ \\ Atunci~e^{x^2} \ge x^2+1.$

$\displaystyle Deci~pentru~x\ \textgreater \ 0:~ \int\limits^x_0 {e^{t^2}} dt \ge \int\limits^x_0(x^2+1) dt =\frac{x^3}{3}+x-1. \\ \\ Rezulta~\lim_{x \to \infty} \int\limits^x_0 {e^{t^2}} dt = +\infty. \\ \\ Iar~\lim_{x \to \infty} \frac{f(x)}{x}=+\infty. \\ \\ Deci~ne~aflam~in~cazul~\frac{\infty}{\infty},~si~deci~putem~aplica~L'Hospital.$

$\displaystyle Limita= \lim_{x \to \infty} \frac{F'(x)}{\left(\frac{f(x)}{x} \right)'}= \lim_{x \to \infty} \frac{f(x)}{\frac{f'(x)x-f(x)}{x^2}}=\lim_{x \to \infty} \frac{x^2}{2x^2-1}= \frac{1}{2}.$