Hasil dari [tex]\displaystyle{ \lim_{x \to 0} \frac{2\sqrt{1+sin5x}-2\sqrt{1-sin5x}}{5x} }[/tex] adalah 2.
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](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{2\sqrt{1+sin5x}-2\sqrt{1-sin5x}}{5x}= }[/tex]
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DITANYA
Tentukan hasilnya.
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PENYELESAIAN
[tex]\displaystyle{ \lim_{x \to 0} \frac{2\sqrt{1+sin5x}-2\sqrt{1-sin5x}}{5x} }[/tex]
[tex]\displaystyle{=\lim_{x \to 0} \frac{2(\sqrt{1+sin5x}-\sqrt{1-sin5x})}{5x}\times\frac{\sqrt{1+sin5x}+\sqrt{1-sin5x}}{\sqrt{1+sin5x}+\sqrt{1-sin5x}} }[/tex]
[tex]\displaystyle{=\frac{2}{5}\lim_{x \to 0} \frac{(\sqrt{1+sin5x})^2-(\sqrt{1-sin5x})^2}{x(\sqrt{1+sin5x}+\sqrt{1-sin5x})} }[/tex]
[tex]\displaystyle{=\frac{2}{5}\lim_{x \to 0} \frac{1+sin5x-1+sin5x}{x(\sqrt{1+sin5x}+\sqrt{1-sin5x})} }[/tex]
[tex]\displaystyle{=\frac{2}{5}\lim_{x \to 0} \frac{2sin5x}{x(\sqrt{1+sin5x}+\sqrt{1-sin5x})} }[/tex]
[tex]\displaystyle{=\frac{4}{5}\lim_{x \to 0} \frac{sin5x}{x}\times\lim_{x \to 0} \frac{1}{\sqrt{1+sin5x}+\sqrt{1-sin5x}} }[/tex]
[tex]\displaystyle{=\frac{4}{5}\times5\times\frac{1}{\sqrt{1+sin5(0)}+\sqrt{1-sin5(0)}} }[/tex]
[tex]\displaystyle{=4\times\frac{1}{\sqrt{1+0}+\sqrt{1-0}} }[/tex]
[tex]\displaystyle{=4\times\frac{1}{2} }[/tex]
[tex]\displaystyle{=2 }[/tex]
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KESIMPULAN
Hasil dari [tex]\displaystyle{ \lim_{x \to 0} \frac{2\sqrt{1+sin5x}-2\sqrt{1-sin5x}}{5x} }[/tex] adalah 2.
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PELAJARI LEBIH LANJUT
- Limit fungsi trigonometri : https://brainly.co.id/tugas/41998117
- Limit trigonometri : https://brainly.co.id/tugas/38915095
- 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
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