Question

Salut, ma puteti ajuta la problema 1230...? Pe mine ma intereseaza cum il aflu pe acel F(x)...

Answer 1

1230.

$f:\mathbb{R}\to \mathbb{R},\quad f(x) = e^{x^2} \\ \\ f(x)-\text{functie continua pe domeniul maxim de }\mathrm{de f initie.} \\ \\ \Rightarrow f(x) - \text{admite primitive pe }\mathbb{R}.\\ \\ \Rightarrow F(x) \text{ are sens in }x = 0 \\ \\\\\lim\limits_{x\to 0}\dfrac{xF(x)}{f(x)} = \lim\limits_{x\to 0}\Big(0\cdot \dfrac{F(0)+C}{f(0)}\Big) = \lim\limits_{x\to 0}\Big(0\cdot \dfrac{F(0)+C}{1}\Big) = 0$

1231.

$\lim\limits_{x\to \infty}\dfrac{x\int e^{x^2} \, dx}{e^{x^2}} \overset{^{\frac{\infty}{\infty}}}{=} \lim\limits_{x\to \infty}\dfrac{\Big(x\int e^{x^2} \, dx\Big)'}{(e^{x^2})'}=\\ \\ = \lim\limits_{x\to \infty}\dfrac{xe^{x^2}+\int e^{x^2}\,dx} {2xe^{x^2}} \overset{^{\frac{\infty}{\infty}}}{=} \lim\limits_{x\to \infty}\dfrac{\Big(xe^{x^2}+\int e^{x^2}\,dx\Big)'} {(2xe^{x^2})'} =$

$=\lim\limits_{x\to \infty}\dfrac{2x^2e^{x^2}+2e^{x^2}}{4x^2e^{x^2}} =\lim\limits_{x\to \infty}\Big(\dfrac{2x^2}{4x^2e^{x^2}}+\dfrac{2e^{x^2}}{4x^2e^{x^2}}\Big) =\\ \\ = \dfrac{1}{2}+\lim\limits_{x\to \infty}\dfrac{1}{2x^2} = \boxed{\dfrac{1}{2}}$

Answer 2

Răspuns:

Explicație pas cu pas: