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math1994
Pe: 2012-06-05T15:41:02+03:00 In: In:

Sa se calculeze lim x->0[ (sin x-tg x)/(x^2*tg x)]

Sa se calculeze lim x->0[ (sin x-tg x)/(x^2*tg x)]

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3 raspunsuri

  1. andrei_93

    Scrie sin in functie de tg.
    \lim_{x\to0}\frac{\frac{2tg\frac{x}{2}}{1+tg^2\frac{x}{2}}-tgx}{x^{2tgx}}=\lim_{x\to0}\frac{\frac{\frac{2tg\frac{x}{2}}{\frac{x}{2}}\frac{x}{2}}{1+\frac{tg^2\frac{x}{2}}{\frac{x^2}{4}}\frac{x^2}{4}}-\frac{tgx}{x}x}{x^{2\frac{tgx}{x}x}}
    Acem limitele remarcabile si rezulta:
    \lim_{x\to0}\frac{\frac{4x-4x-x^3}{4+x^2}}{x^{2x}}=\lim_{x\to0}\frac{\frac{-x^3}{4+x^2}}{e^{2xlnx}}=\lim_{x\to0}\frac{\frac{-x^3}{4+x^2}}{e^\frac{lnx}{2x^{-1}}}=\lim_{x\to0}\frac{\frac{-x^3}{4+x^2}}{e^\frac{lnx}{2x^{-1}}}=\lim_{x\to0}\frac{\frac{-x^3}{4+x^2}}{e^0}=0

  2. DD
    profesor

    Cred ca limita este ; L=lim(x->0)[(sin(x)-tg(x))/((x^2).(tg(x))]=lim(x->0)[
    (sin(x)/x).(1-1/cos(x)) / ((x^2).(tg(x)/x))]=lim(x->0)[(sin(x)/x)] . lim(x
    ->0)[1/cos(x)] . lim(x->0)[1/(tg(x)/x)] . lim(x->0)[((cos(x))-1)/x^2]=1.lim(x->0)[(-2.(sin(x/2))^2)/x^2]=-2.(1/2)^2=-1/2. [Am folosit formulele ;
    2.(sin(a))^2=1-cos(2.a) , pentru a=x/2 , lim(x->0)[(sin(x))/x]=1 , lim(x->0)[(tg(x))/x]=1 , lim(x->0)[cos(x)]=1]

  4. edy8
    maestru (V)

    \it{\frac{sinx - tgx}{x^2tgx} = \frac{1}{x^2}(\frac{sinx}{tgx} - \frac{tgx}{tgx}) = \frac{1}{x^2}(sinx\cdot\frac{1}{tgx} - 1) = \frac{1}{x^2}(sinx\cdot ctgx - 1) = \frac{1}{x^2}(sinx\cdot\frac{cosx}{sinx} - 1) = \frac{1}{x^2}(cosx - 1)\bl\\\;\\Limita ceruta este:\\\;\\\lim_{x\to0} \frac{cosx - 1}{x^2} = l\\\;\\Se aplica, de doua ori, regula L'Hospital pentru cazul \frac{0}{0} si \Rightarrow l = - \frac{1}{2}.}

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