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Berikut ini adalah pertanyaan dari xsx12 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

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pakee cara ya
jan ngasal
bantuupakee cara yajan ngasal

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Nilai dari $\displaystyle{ \lim_{x \to 0} \frac{xtan2x-2xsinx}{(1-cos2x)^2}}$ adalah $\displaystyle{\boldsymbol{\frac{3}{4}}}$ .

PEMBAHASAN

Teorema pada limit adalah sebagai berikut :

$(i)~\lim\limits_{x \to c} f(x)=f(c)$

$(ii)~\lim\limits_{x \to c} kf(x)=k\lim\limits_{x \to c} f(x)$

$(iii)~\lim\limits_{x \to c} [f(x)\pm g(x)]=\lim\limits_{x \to c} f(x)\pm\lim\limits_{x \to c} g(x)$

$(iv)~\lim\limits_{x \to c} [f(x)\times g(x)]=\lim\limits_{x \to c} f(x)\times\lim\limits_{x \to c} g(x)$

$\displaystyle{(v)~\lim\limits_{x \to c} \left [ \frac{f(x)}{g(x)} \right ]=\frac{\lim\limits_{x \to c} f(x)}{\lim\limits_{x \to c} g(x)} }$

$(vi)~\lim\limits_{x \to c} \left [ f(x) \right ]^n=\left [ \lim\limits_{x \to c} f(x) \right ]^n$

Rumus untuk limit fungsi trigonometri :

$\displaystyle{(i)~\lim\limits_{x \to 0} \frac{sinax}{bx}=\lim\limits_{x \to 0} \frac{tanax}{bx}=\frac{a}{b} }$

$\displaystyle{(ii)~\lim\limits_{x \to 0} \frac{ax}{sinbx}=\lim\limits_{x \to 0} \frac{ax}{tanbx}=\frac{a}{b} }$

$\displaystyle{(iii)~\lim\limits_{x \to 0} \frac{sinax}{sinbx}=\lim\limits_{x \to 0} \frac{tanax}{tanbx}=\frac{a}{b} }$

$\displaystyle{(iv)~\lim\limits_{x \to a} \frac{sin(x-a)}{(x-a)}=\lim\limits_{x \to a} \frac{tan(x-a)}{(x-a)}=1 }$

DIKETAHUI

$\displaystyle{ \lim_{x \to 0} \frac{xtan2x-2xsinx}{(1-cos2x)^2}= }$

DITANYA

Tentukan nilainya.

PENYELESAIAN

$\displaystyle{ \lim_{x \to 0} \frac{xtan2x-2xsinx}{(1-cos2x)^2}}$

$\displaystyle{=\lim_{x \to 0} \frac{xtan2x-2xsinx}{[1-(1-2sin^2x)]^2}}$

$\displaystyle{=\lim_{x \to 0} \frac{xtan2x-2xsinx}{(2sin^2x)^2}}$

$\displaystyle{=\lim_{x \to 0} \frac{\frac{xsin2x}{cos2x}-2xsinx}{4sin^4x}}$

$\displaystyle{=\frac{1}{4}\lim_{x \to 0} \frac{\frac{xsin2x-2xsinxcos2x}{cos2x}}{sin^4x}}$

$\displaystyle{=\frac{1}{4}\lim_{x \to 0} \frac{xsin2x-2xsinxcos2x}{sin^4xcos2x}}$

$\displaystyle{=\frac{1}{4}\lim_{x \to 0} \frac{2xsinxcosx-2xsinxcos2x}{sin^4xcos2x}}$

$\displaystyle{=\frac{1}{4}\lim_{x \to 0} \frac{2xsinx(cosx-cos2x)}{sin^4xcos2x}}$

$\displaystyle{=\frac{1}{2}\lim_{x \to 0} \frac{x\left [ -2sin\left ( \frac{x+2x}{2} \right )sin\left ( \frac{x-2x}{2} \right ) \right ]}{sin^3xcos2x}}$

$\displaystyle{=-\lim_{x \to 0} \frac{xsin\left ( \frac{3x}{2} \right )sin\left ( \frac{-x}{2} \right )}{sin^3xcos2x}}$

$\displaystyle{=-\lim_{x \to 0} \frac{-xsin\left ( \frac{3x}{2} \right )sin\left ( \frac{x}{2} \right )}{sin^3xcos2x}}$

$\displaystyle{=\lim_{x \to 0} \frac{xsin\left ( \frac{3x}{2} \right )sin\left ( \frac{x}{2} \right )}{sin^3xcos2x}}$

$\displaystyle{=\lim_{x \to 0} \frac{x}{sinx}\times\lim_{x \to 0} \frac{sin\left ( \frac{3x}{2} \right )}{sinx}\times\lim_{x \to 0} \frac{sin\left ( \frac{x}{2} \right )}{sinx}\times\lim_{x \to 0} \frac{1}{cos2x}}$