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TEORIE

Data publicarii: 24.05.2014

Cazul I: 0/0. 

Fie functiile f,g:I -> R, unde I este interval in R si a este un punct de acumulare al

acestuia. Daca: 

1)\;\lim_{{x}\rightarrow{a}}{f(x)}=\lim_{{x}\rightarrow{a}}{g(x)}=0,1)\;\lim_{{x}\rightarrow{a}}{f(x)}=\lim_{{x}\rightarrow{a}}{g(x)}=0,

2)\;f\;si\;g\;sunt\; derivabile\; pe2)\;f\;si\;g\;sunt\; derivabile\; pe I\setminus{\begin{Bmatrix}{a}\end{Bmatrix}},I\setminus{\begin{Bmatrix}{a}\end{Bmatrix}},

3)\;g(x)\neq{0}\;\forall{x}\in{{I}\setminus{\begin{Bmatrix}{a}\end{Bmatrix}}},3)\;g(x)\neq{0}\;\forall{x}\in{{I}\setminus{\begin{Bmatrix}{a}\end{Bmatrix}}},

4)\;{{{g}^{ 4)\;{{{g}^{ '}}(x)}\neq{0},\forall{x}\in{{I}\setminus{\begin{Bmatrix}{a}\end{Bmatrix}}},

5)\;\exists\lim_{{x}\rightarrow{a}}{\frac{{{f}^{5)\;\exists\lim_{{x}\rightarrow{a}}{\frac{{{f}^{'}}(x)}{{{g}^{'}}(x)}\in{\bar{\mathbb{R}}}},

atunci functia f/g are limita in x = a si:

CONTINUARE LA : TEORIE

EXERCITIUL 10

Data publicarii: 05.11.2014

Suport teoretic:

Limite de functii,integrale definite,regula L'Hospital,arctangenta,logaritm natural.

Enunt:

Fie functiile:

f,g:[e,+oo) - > R,

f(t) = arctg(lnt),

g(t) = ln(arctgt).

Sa se calculeze: 

L=lim_{x\searrow{e}}{\frac{\int_e^x{f(t)dt}}{\int_e^x{g(t)dt}}}.L=lim_{x\searrow{e}}{\frac{\int_e^x{f(t)dt}}{\int_e^x{g(t)dt}}}.

Raspuns:

L=\frac{\pi}{4ln[(arctg(e)]}.L=\frac{\pi}{4ln[(arctg(e)]}.

CONTINUARE LA : EXERCITIUL 10

EXERCITIUL 9

Data publicarii: 04.11.2014

Suport teoretic:

Limite de functii,regulile lui l'Hospital.

Enunt:

Sa se calculeze:

L=\lim_{x\rightarrow{2}}{\frac{{x^4}-5{x^3}+6{x^2} +4x-8}{{x^4}-7{x^3} +18{x^2}- 20x+8}}.L=\lim_{x\rightarrow{2}}{\frac{{x^4}-5{x^3}+6{x^2} +4x-8}{{x^4}-7{x^3} +18{x^2}- 20x+8}}.

Raspuns: 

L = 3. 

CONTINUARE LA : EXERCITIUL 9

EXERCITIUL 8

Data publicarii: 04.11.2014

Suport teoretic:

Prelungire prin continuitate,limite laterale,regula lui l'Hôspital.

Enunt:

Determinati prelungirea prin continuitate, in punctul x = π/2, a functiei:

f:{(0,\frac{\pi}{2})}\rightarrow{\mathbb{R}},\;f(x)={(\sin{x})}^{\ln{(tgx)}}.f:{(0,\frac{\pi}{2})}\rightarrow{\mathbb{R}},\;f(x)={(\sin{x})}^{\ln{(tgx)}}.

Réponse: 

\tilde{f}:{(0,\frac{\pi}{2}]}\rightarrow{\mathbb{R}},\tilde{f}(x)=\begin{cases}f(x),x\in(0,\frac{\pi}{2})\\1,x=\frac{\pi}{2}\end{cases}.\tilde{f}:{(0,\frac{\pi}{2}]}\rightarrow{\mathbb{R}},\tilde{f}(x)=\begin{cases}f(x),x\in(0,\frac{\pi}{2})\\1,x=\frac{\pi}{2}\end{cases}.

CONTINUARE LA : EXERCITIUL 8

EXERCITIUL 7

Data publicarii: 02.11.2014

Suport teoretic:

Limite siruri,limite functii,integrale definite,primitive,definitia derivatei,regula lui l'Hospital.

Enunt:

Sa se calculeze:

L=\lim_{x\rightarrow{\infty}}[\lim_{n\rightarrow{\infty}}L=\lim_{x\rightarrow{\infty}}[\lim_{n\rightarrow{\infty}} ({n}\cdot{\int_{-x}^{x+\frac{1}{n}}{\frac{t}{e^{t^2}}{dt}})].}({n}\cdot{\int_{-x}^{x+\frac{1}{n}}{\frac{t}{e^{t^2}}{dt}})].}

Raspuns:

L = 0.

CONTINUARE LA : EXERCITIUL 7

 

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