Informaţii, definiţii, teoreme, formule, exerciţii şi probleme rezolvate din matematica de liceu. RSS/XML

»LIMITE DE FUNCTII

Data publicarii: 26 Octombrie, 2008

TEORIE

Definitia limitei unei functii intr-un punct (definitia lui Heine):

Fie a un punct de acumulare (finit sau infinit) al unei mulţimi E.

Se spune că L (din R, sau +/-00) este limita funcţiei f:E --> R in punctul

a, daca oricare ar fi xn din E, xn diferit de a, pentru orice n natural, xn - > a,

sirul (f(xn)), al valorilor functiei, tinde catre L (din R, sau +/-00).

Teorema clestelui (t eorema celor doi jandarmi):

Fie 3 functii f,g,h:E -> R, a un punct de acumulare pentru E si V o vecinatate a lui a.

Daca:

$a)\;{f(x)}\leq{g(x)}\leq{h(x)},\forall{x}\in{{\mathcal{V}}\cap{E}},x\not=a\;si$ a)\;{f(x)}\leq{g(x)}\leq{h(x)},\forall{x}\in{{\mathcal{V}}\cap{E}},x\not=a\;si

$b)\;{\lim}_{{x}\rightarrow{a}}{f(x)}={\lim}_{{x}\rightarrow{a}}{h(x)}=L,$ b)\;{\lim}_{{x}\rightarrow{a}}{f(x)}={\lim}_{{x}\rightarrow{a}}{h(x)}=L,

atunci g are limita in a si:

${\lim}_{{x}\rightarrow{a}}{g(x)} =L.$ {\lim}_{{x}\rightarrow{a}}{g(x)} =L.

Limite remarcabile:

$\lim_{{x}\rightarrow{0}}\frac{\sin{x}}{x} =1.$ \lim_{{x}\rightarrow{0}}\frac{\sin{x}}{x} =1.
$\lim_{{x}\rightarrow{a}}\frac{\sin{u(x)}}{u(x)} =1,\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.$ \lim_{{x}\rightarrow{a}}\frac{\sin{u(x)}}{u(x)} =1,\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.
$\lim_{{x}\rightarrow{0}}\frac{tgx}{x} =1.$ \lim_{{x}\rightarrow{0}}\frac{tgx}{x} =1.
$\lim_{{x}\rightarrow{a}}\frac{{tg}{u(x)}}{u(x)} =1,\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.$ \lim_{{x}\rightarrow{a}}\frac{{tg}{u(x)}}{u(x)} =1,\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.

$\lim_{{x}\rightarrow{0}}\frac{\arcsin{x}}{x} =1.$ \lim_{{x}\rightarrow{0}}\frac{\arcsin{x}}{x} =1.
$\lim_{{x}\rightarrow{0}}\frac{arctgx}{x} =1.$ \lim_{{x}\rightarrow{0}}\frac{arctgx}{x} =1.
$\lim_{{x}\rightarrow{a}}\frac{\arcsin{u(x)}}{u(x)} =1,\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.$ \lim_{{x}\rightarrow{a}}\frac{\arcsin{u(x)}}{u(x)} =1,\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.
$\lim_{{x}\rightarrow{a}}\frac{{arctg}{u(x)}}{u(x)} =1,\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.$ \lim_{{x}\rightarrow{a}}\frac{{arctg}{u(x)}}{u(x)} =1,\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.
$\lim_{{x}\rightarrow{\pm\infty}}{(1+\frac{1}{x})}^{x}=e.$ \lim_{{x}\rightarrow{\pm\infty}}{(1+\frac{1}{x})}^{x}=e.
$\lim_{{x}\rightarrow{0}}{(1+x)}^{\frac{1}{x}}={e}.$ \lim_{{x}\rightarrow{0}}{(1+x)}^{\frac{1}{x}}={e}.
$\lim_{{x}\rightarrow{a}}{(1+\frac{1}{u(x)})}^{u(x)}={e},\;daca\; \lim_{{x}\rightarrow{a}}{u(x)}={\pm\infty}.$ \lim_{{x}\rightarrow{a}}{(1+\frac{1}{u(x)})}^{u(x)}={e},\;daca\; \lim_{{x}\rightarrow{a}}{u(x)}={\pm\infty}.
$\lim_{{x}\rightarrow{a}}{(1+u(x))}^{\frac{1}{u(x)}}={e},\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.$ \lim_{{x}\rightarrow{a}}{(1+u(x))}^{\frac{1}{u(x)}}={e},\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.
$\lim_{{x}\rightarrow{0}}\frac{\ln{(1+x)}}{x}={1}.$ \lim_{{x}\rightarrow{0}}\frac{\ln{(1+x)}}{x}={1}.
$\lim_{{x}\rightarrow{a}}\frac{\ln{(1+u(x))}}{u(x)}={1},\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.$ \lim_{{x}\rightarrow{a}}\frac{\ln{(1+u(x))}}{u(x)}={1},\;daca\;\lim_{{x}\rightarrow{a}}{u(x)}=0.
$\lim_{{x}\rightarrow{0}}\frac{a^x-1}{x}=\ln{a}, a > 0.$ \lim_{{x}\rightarrow{0}}\frac{a^x-1}{x}=\ln{a}, a > 0.
$\lim_{{x}\rightarrow{0}}\frac{e^x-1}{x}={1}$ \lim_{{x}\rightarrow{0}}\frac{e^x-1}{x}={1} .
$\lim_{{x}\rightarrow{\alpha}}\frac{a^{u(x)}-1}{u(x)}=\ln{a}, a> 0,\;daca\;\lim_{{x}\rightarrow{\alpha}}{u(x)}=0.$ \lim_{{x}\rightarrow{\alpha}}\frac{a^{u(x)}-1}{u(x)}=\ln{a}, a> 0,\;daca\;\lim_{{x}\rightarrow{\alpha}}{u(x)}=0.
$\lim_{{x}\rightarrow{0}}\frac{{(1+x)}^{r}-1}{x}={r},\; unde\; r\in{\mathbb{R}}.$ \lim_{{x}\rightarrow{0}}\frac{{(1+x)}^{r}-1}{x}={r},\; unde\; r\in{\mathbb{R}}.
$\lim_{{x}\rightarrow{\alpha}}\frac{e^{u(x)}-1}{u(x)}={1},\;daca\;\lim_{{x}\rightarrow{\alpha}}{u(x)}=0.$ \lim_{{x}\rightarrow{\alpha}}\frac{e^{u(x)}-1}{u(x)}={1},\;daca\;\lim_{{x}\rightarrow{\alpha}}{u(x)}=0.

Postat în LIMITE DE FUNCTII

Revino la categoria principală

Adăugaţi un comentariu

Răspunsuri şi comentarii

Până acum, niciun comentariu nu a fost adăugat.

Selecteaza acest link pentru a ma contacta prin YAHOO MESSENGER !

Trimite un sms şi beneficiezi de consultanţă online !

CATEGORII :

Arhiva blog-ului

Developed by Hagau Ioan