Question

Nilai dari lim x -> 0 (1 - cos^3x)/(sin^2 x) adalah....
a. -3/2
b. -1/2
c. 0
d. 1/2
e. 3/2​

Answer 1

Hasil dari [tex]\displaystyle{ \lim_{x \to 0} \frac{1-cos^3x}{sin^2x} }[/tex] adalah [tex]\displaystyle{\boldsymbol{e.~\frac{3}{2}}}[/tex].

PEMBAHASAN

Teorema pada limit adalah sebagai berikut :

[tex](i)~\lim\limits_{x \to c} f(x)=f(c)[/tex]

[tex](ii)~\lim\limits_{x \to c} kf(x)=k\lim\limits_{x \to c} f(x)[/tex]

[tex](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)[/tex]

[tex](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)[/tex]

[tex]\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)} }[/tex]

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

Rumus untuk limit fungsi trigonometri :

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

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

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

[tex]\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 }[/tex]

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DIKETAHUI

[tex]\displaystyle{ \lim_{x \to 0} \frac{1-cos^3x}{sin^2x}= }[/tex]

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DITANYA

Tentukan nilai limitnya.

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PENYELESAIAN

[tex]\displaystyle{ \lim_{x \to 0} \frac{1-cos^3x}{sin^2x} }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \frac{1-(cos^2x)cosx}{sin^2x} }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \frac{1-(1-sin^2x)cosx}{sin^2x} }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \frac{1-cosx+sin^2xcosx}{sin^2x} }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \left ( \frac{1-cosx}{sin^2x}+\frac{sin^2xcosx}{sin^2x} \right ) }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \left ( \frac{1-cosx}{sin^2x}\times\frac{1+cosx}{1+cosx}+cosx \right ) }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \left ( \frac{1-cos^2x}{sin^2x(1+cosx)}+cosx \right ) }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \left ( \frac{sin^2x}{sin^2x(1+cosx)}+cosx \right ) }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \left ( \frac{1}{1+cosx}+cosx \right ) }[/tex]

[tex]\displaystyle{=\frac{1}{1+cos0}+cos0 }[/tex]

[tex]\displaystyle{=\frac{1}{1+1}+1 }[/tex]

[tex]\displaystyle{=\frac{1}{2}+1 }[/tex]

[tex]\displaystyle{=\frac{3}{2}}[/tex]

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> Menggunakan Aturan L'hospital :

[tex]\displaystyle{ \lim_{x \to 0} \frac{1-cos^3x}{sin^2x} }[/tex]

[tex]\displaystyle{=\frac{1-cos^30}{sin^20} }[/tex]

[tex]\displaystyle{=\frac{0}{0} }[/tex]

Karena substitusi langsung hasilnya bentuk tak tentu, bisa kita gunakan aturan l'hospital.

[tex]\displaystyle{ \lim_{x \to 0} \frac{1-cos^3x}{sin^2x} }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \frac{\frac{d}{dx}\left ( 1-cos^3x \right )}{\frac{d}{dx}\left ( sin^2x \right )} }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \frac{-3cos^2x(-sinx)}{2sinxcosx} }[/tex]

[tex]\displaystyle{=\lim_{x \to 0} \frac{3cosx}{2} }[/tex]

[tex]\displaystyle{=\frac{3cos0}{2} }[/tex]

[tex]\displaystyle{=\frac{3}{2} }[/tex]

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KESIMPULAN

Hasil dari [tex]\displaystyle{ \lim_{x \to 0} \frac{1-cos^3x}{sin^2x} }[/tex] adalah [tex]\displaystyle{\boldsymbol{e.~\frac{3}{2}}}[/tex].

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PELAJARI LEBIH LANJUT

1. Limit trigonometri : https://brainly.co.id/tugas/41998117
2. Limit trigonometri : https://brainly.co.id/tugas/38915095
3. Limit trigonometri : https://brainly.co.id/tugas/30308496

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DETAIL JAWABAN

Kelas : 11

Mapel: Matematika

Bab : Limit Fungsi

Kode Kategorisasi: 11.2.8