Berapa hasil dari limit ini
[tex]\displaystyle \lim_{x\to 0}\frac{x^3}{\sqrt{1+\tan x}-\sqrt{1+\sin x}}[/tex]
[tex]\displaystyle \lim_{x\to 0}\frac{x^3}{\sqrt{1+\tan x}-\sqrt{1+\sin x}}[/tex]
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Hasil dari [tex]\displaystyle{ \lim_{x \to 0} \frac{x^3}{\sqrt{1+tanx}-\sqrt{1+sinx}} }[/tex] adalah 4.
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{x^3}{\sqrt{1+tanx}-\sqrt{1+sinx}}= }[/tex]
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DITANYA
Tentukan hasilnya.
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PENYELESAIAN
Kalikan dengan akar sekawannya.
[tex]\displaystyle{ \lim_{x \to 0} \frac{x^3}{\sqrt{1+tanx}-\sqrt{1+sinx}} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3}{\sqrt{1+tanx}-\sqrt{1+sinx}}\times\frac{\sqrt{1+tanx}+\sqrt{1+sinx}}{\sqrt{1+tanx}+\sqrt{1+sinx}} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3(\sqrt{1+tanx}+\sqrt{1+sinx})}{1+tanx-(1+sinx)} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3(\sqrt{1+tanx}+\sqrt{1+sinx})}{tanx-sinx} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3(\sqrt{1+tanx}+\sqrt{1+sinx})}{\frac{sinx}{cosx}-sinx} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3(\sqrt{1+tanx}+\sqrt{1+sinx})}{\frac{sinx-sinxcosx}{cosx}} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3cosx(\sqrt{1+tanx}+\sqrt{1+sinx})}{sinx-sinxcosx} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3cosx(\sqrt{1+tanx}+\sqrt{1+sinx})}{sinx(1-cosx)}\times\frac{1+cosx}{1+cosx} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3cosx(1+cosx)(\sqrt{1+tanx}+\sqrt{1+sinx})}{sinx(1-cos^2x)} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3cosx(1+cosx)(\sqrt{1+tanx}+\sqrt{1+sinx})}{sinx(sin^2x)} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{x^3cosx(1+cosx)(\sqrt{1+tanx}+\sqrt{1+sinx})}{sin^3x} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \left ( \frac{x}{sinx} \right )^3\times\lim_{x \to 0} (cosx+cos^2x)\times\lim_{x \to 0} (\sqrt{1+tanx}+\sqrt{1+sinx}) }[/tex]
[tex]\displaystyle{=\left ( 1 \right )^3\times(cos0+cos^20)\times(\sqrt{1+tan0}+\sqrt{1+sin0}) }[/tex]
[tex]\displaystyle{=(1+1)\times(\sqrt{1+0}+\sqrt{1+0}) }[/tex]
[tex]\displaystyle{=2\times2 }[/tex]
[tex]=4[/tex]
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KESIMPULAN
Hasil dari [tex]\displaystyle{ \lim_{x \to 0} \frac{x^3}{\sqrt{1+tanx}-\sqrt{1+sinx}} }[/tex] adalah 4.
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PELAJARI LEBIH LANJUT
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DETAIL JAWABAN
Kelas : 11
Mapel: Matematika
Bab : Limit Fungsi
Kode Kategorisasi: 11.2.8
Verified answer
PEMBAHASAN
lim x→0 ax/sin bx = a/b
lim x→0 x/sin x = 1
(a + b)(a - b) = a² - b²
cos 2x = 1 - 2 sin² x
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