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Data publicarii: 31 Ianuarie, 2010

ANALIZĂ-38

Suport teoretic:

Functie multiforma, limite laterale, functii continue, cazuri exceptate la limite de functii, regula lui l'Hospital.

Enunt:

Sa se studieze continuitatea functiei f:(1,+oo) - > R, unde

$f(x)=\begin{cases}{(1+{cos}{\frac{\pi}{x}})}^{{tg}{\frac{\pi}{x}}},\;{x}\in{(1,2)}\\e,\;x=2\\{(x-1)}^{\frac{1}{x-2}},\;{x}\in{(2,\infty)}\end{cases}.$ f(x)=\begin{cases}{(1+{cos}{\frac{\pi}{x}})}^{{tg}{\frac{\pi}{x}}},\;{x}\in{(1,2)}\\e,\;x=2\\{(x-1)}^{\frac{1}{x-2}},\;{x}\in{(2,\infty)}\end{cases}.

Raspuns:

Functia f este continua pe domeniul sau de definitie.

Rezolvare:

Se calculeaza limitele laterale in x = 2 si se compara cu valoarea functiei in x = 2:

${f_s}(2)=\lim_{x\nearrow{2}}{(1+\cos{\frac{\pi}{x}})}^{{tg}{\frac{\pi}{x}}}=(1^{\infty})=$ {f_s}(2)=\lim_{x\nearrow{2}}{(1+\cos{\frac{\pi}{x}})}^{{tg}{\frac{\pi}{x}}}=(1^{\infty})= $e^{\lim_{x\nearrow{2}}{{tg}{\frac{\pi}{x}}\cdot{{ln}(1+{cos}{\frac{\pi}{x}})}}}=$ e^{\lim_{x\nearrow{2}}{{tg}{\frac{\pi}{x}}\cdot{{ln}(1+{cos}{\frac{\pi}{x}})}}}=

${e}^{\lim_{x\nearrow{2}}{\frac{{ln}{(1+{cos}{\frac{\pi}{x}})}}{{ctg}{\frac{\pi}{x}}}}}=$ {e}^{\lim_{x\nearrow{2}}{\frac{{ln}{(1+{cos}{\frac{\pi}{x}})}}{{ctg}{\frac{\pi}{x}}}}}= $(\frac{0}{0},\;l$ (\frac{0}{0},\;l'Hospital)=...=e,\;(1)

${f_d}(2)=\lim_{x\searrow{2}}{{(x-1)}^{\frac{1}{x-2}}}=$ {f_d}(2)=\lim_{x\searrow{2}}{{(x-1)}^{\frac{1}{x-2}}}= $(1^{\infty})=$ (1^{\infty})= $e^{\lim_{x\searrow{2}}{{\frac{1}{x-2}}\cdot{{ln}(x-1)}}}=$ e^{\lim_{x\searrow{2}}{{\frac{1}{x-2}}\cdot{{ln}(x-1)}}}= $(\frac{0}{0},\;l$ (\frac{0}{0},\;l'Hospital)=...=e,\;(2)
$f(2)=e.\;(3)$ f(2)=e.\;(3)