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maestru (V)
Pe: 2014-03-27T15:21:51+02:00 In: In:

limita

lim(x-pi/3) (sin(x/2)-1/2 )/(x-pi/3) eu obtin 1/2 ;l-am scris pe 1/2 ca sin pi/6 si am transformat acea suma in produs.Daca fac cu l`hospital obtin 1/4.Unde gresesc?

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  1. expert (VI)

    Limita este derivata lui \sin \frac{x}{2} in \frac{\pi}{3}. Avem f(x)=\sin \frac{x}{2} \Rightarrow f'(x)=\cos \frac{x}{2}\cdot \left( \frac{x}{2}\right)'=\frac{1}{2}\cos \frac{x}{2}, deci limita e f'\left( \frac{\pi}{3}\right) = \frac{1}{2}\cdot \frac{\sqrt 3}{2}=\frac{\sqrt 3}{4}.

    Cu , pe care o putem aplica caci suntem in cazul de nedeterminare \frac{0}{0}, limita ceruta este \lim_{x\to \frac{\pi}{3}} \left( \sin \frac{x}{2} \right)' =\lim_{x\to\frac{\pi}{3}} \frac{1}{2}\cos \frac{x}{2} = \frac{\sqrt 3}{4}.

    Cu transformare in produs:
    \sin \frac{x}{2}- \sin \frac{\pi}{6} = 2\cos \frac{\frac{x}{2}+\frac{\pi}{6}}{2}\sin \frac{\frac{x}{2}-\frac{\pi}{6}}{2}=2\cos \frac{3x+\pi}{12}\sin \frac{x-\frac{\pi}{3}}{4}. Acum limita este egala cu \lim_{x\to\frac{\pi}{3}}\frac{2\cos \frac{3x+\pi}{12}\sin \frac{x-\frac{\pi}{3}}{4}}{x-\frac{\pi}{3}}=2\lim_{x\to\frac{\pi}{3}}\cos \frac{3x+\pi}{12}\cdot \lim_{x\to\frac{\pi}{3}} \frac{1}{4}\cdot \frac{\sin \frac{x-\frac{\pi}{3}}{4}}{\frac{x-\frac{\pi}{3}}{4}}=2\cos \frac{\pi}{6} \cdot \frac{1}{4}\cdot 1=\frac{\sqrt 3}{2}\cdot \frac{1}{2}=\frac{\sqrt 3}{4}.

    Va multumesc pentru ideile de abordare!:D

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  2. maestru (V)

    Da multumesc,am gresit ca un prost la calcule,am uitat acel 2 din formula 🙁!

    • 0
  4. expert (VI)

    Cu multa placere!:D Stati linistit, se mai intampla.. ^_^

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