Matematika

Pertanyaan

lim x->pi 1-sin^2x/(sin 1/2x - cos 1/2x)^2

1 Jawaban

  • [tex]1-\sin^{2}x=(1-\sin(x))(1+\sin(x))[/tex]
    [tex](\sin(\frac{1}{2}x)-\cos(\frac{1}{2}x)) ^{2}=\sin^{2}(\frac{1}{2}x)-2\sin(\frac{1}{2}x) \cos(\frac{1}{2}x) + \cos^{2}(\frac{1}{2}x)))[/tex]
    jadi
    [tex]\lim\limits_{x \rightarrow \pi}\frac{1-\sin^{2}x}{(\sin(\frac{1}{2}x)-\cos(\frac{1}{2}x))^{2}}[/tex]
    [tex]\lim\limits_{x \rightarrow \pi}\frac{(1-\sin(x))(1+\sin(x))}{1-\sin(x)}[/tex]
    [tex]\lim\limits_{x \rightarrow \pi}1+\sin(x)[/tex]
    [tex]=1+\sin(\pi)=0/tex]

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