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Data publicarii: 04 Noiembrie, 2014

EXERCITIUL 8

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}.

Rezolvare:

Conform definitiei, se impune:

f(\frac{\pi}{2})=\lim_{x\nearrow{\frac{\pi}{2}}}f(\frac{\pi}{2})=\lim_{x\nearrow{\frac{\pi}{2}}} f(x)=\lim_{x\nearrow{\frac{\pi}{2}}}f(x)=\lim_{x\nearrow{\frac{\pi}{2}}} (\sin{x})^{\ln{(tgx)}}=(\sin{x})^{\ln{(tgx)}}=

=({1}^{\infty})==({1}^{\infty})= \lim_{x\nearrow{\frac{\pi}{2}}}\lim_{x\nearrow{\frac{\pi}{2}}} {e}^{{\ln{(tgx)}}{\ln{(\sin{x})}}}={e}^{{\ln{(tgx)}}{\ln{(\sin{x})}}}=

={e}^{\lim_{x\nearrow{\frac{\pi}{2}}}[ln(tgx)][ln(sinx)]}=={e}^{\lim_{x\nearrow{\frac{\pi}{2}}}[ln(tgx)][ln(sinx)]}= ({0} \cdot{\infty})=({0} \cdot{\infty})= e^{\mathcal{L}},e^{\mathcal{L}},

unde\;\mathcal{L}=\lim_{x\nearrow{\frac{\pi}{2}}}\frac{\ln{(\sin{x})}}{\frac{1}{\ln{(tgx)}}}=unde\;\mathcal{L}=\lim_{x\nearrow{\frac{\pi}{2}}}\frac{\ln{(\sin{x})}}{\frac{1}{\ln{(tgx)}}}= (\frac{0}{0});(\frac{0}{0});

se aplica regula lui l'Hôspital si:

\mathcal{L}=\lim_{x\nearrow{\frac{\pi}{2}}}\frac{[\ln{(\sin{x})]\mathcal{L}=\lim_{x\nearrow{\frac{\pi}{2}}}\frac{[\ln{(\sin{x})]'}}{[\frac{1}{\ln{(tgx)}}]'} =\cdots==\cdots= \lim_{t\rightarrow{\infty}}\frac{{(\ln{t})}^2}{1+{t}^{2}},\lim_{t\rightarrow{\infty}}\frac{{(\ln{t})}^2}{1+{t}^{2}}, unde\;t={tgx}.unde\;t={tgx}.

Se aplica inca de 2 ori regula lui l'Hôspital si se obtine

\mathcal{L}=0,\,{deci}\,f(\frac{\pi}{2})=e^0=1,\mathcal{L}=0,\,{deci}\,f(\frac{\pi}{2})=e^0=1,

de unde rezulta solutia.


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