👤
RazvanInfo
a fost răspuns

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



Salut Ma Puteti Ajuta La Problema 1230 Pe Mine Ma Intereseaza Cum Il Aflu Pe Acel Fx class=

Răspuns :

Rayzen

1230.

[tex]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[/tex]

1231.

[tex]\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})'} =[/tex]

[tex]=\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}}[/tex]

Răspuns:

Explicație pas cu pas:

Vezi imaginea Halogenhalogen