toloongg dong bantuiinn
1. Hasil dari [tex]\displaystyle{ \lim_{t \to 2} \frac{(t^2-5t+6)sin(t-2)}{(t^2-t-2)^2}}[/tex] adalah [tex]\displaystyle{\boldsymbol{-\frac{1}{9} }}[/tex].
2. Hasil dari [tex]\displaystyle{\lim_{x \to -2} \frac{1-cos(x+2)}{x^2+4x+4}}[/tex] adalah [tex]\displaystyle{\boldsymbol{\frac{1}{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{1.~ \lim_{t \to 2} \frac{(t^2-5t+6)sin(t-2)}{(t^2-t-2)^2}= }[/tex]
[tex]\displaystyle{2.~ \lim_{x \to -2} \frac{1-cos(x+2)}{x^2+4x+4}= }[/tex]
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DITANYA
Tentukan hasilnya.
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PENYELESAIAN
Soal 1.
[tex]\displaystyle{\lim_{t \to 2} \frac{(t^2-5t+6)sin(t-2)}{(t^2-t-2)^2} }[/tex]
[tex]\displaystyle{=\lim_{t \to 2} \frac{(t-2)(t-3)sin(t-2)}{[(t-2)(t+1)]^2} }[/tex]
[tex]\displaystyle{=\lim_{t \to 2} \frac{(t-2)(t-3)sin(t-2)}{(t-2)^2(t+1)^2} }[/tex]
[tex]\displaystyle{=\lim_{t \to 2} \frac{(t-3)sin(t-2)}{(t-2)(t+1)^2} }[/tex]
[tex]\displaystyle{=\lim_{t \to 2} \frac{(t-3)}{(t+1)^2}\times\lim_{t \to 2} \frac{sin(t-2)}{t-2} }[/tex]
[tex]\displaystyle{=\frac{(2-3)}{(2+1)^2}\times1 }[/tex]
[tex]\displaystyle{=-\frac{1}{9} }[/tex]
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Soal 2.
[tex]\displaystyle{\lim_{x \to -2} \frac{1-cos(x+2)}{x^2+4x+4} }[/tex]
[tex]\displaystyle{=\lim_{x \to -2} \frac{1-cos(x+2)}{(x+2)^2}\times\frac{1+cos(x+2)}{1+cos(x+2)} }[/tex]
[tex]\displaystyle{=\lim_{x \to -2} \frac{1-cos^2(x+2)}{(x+2)^2[1+cos(x+2)]} }[/tex]
[tex]\displaystyle{=\lim_{x \to -2} \frac{sin^2(x+2)}{(x+2)^2[1+cos(x+2)]} }[/tex]
[tex]\displaystyle{=\lim_{x \to -2} \left [ \frac{sin(x+2)}{(x+2)} \right ]^2\times\lim_{x \to -2} \frac{1}{1+cos(x+2)} }[/tex]
[tex]\displaystyle{= \left [ \lim_{x \to -2} \frac{sin(x+2)}{(x+2)} \right ]^2\times\frac{1}{1+cos(-2+2)} }[/tex]
[tex]\displaystyle{= 1^2\times\frac{1}{1+cos(0)} }[/tex]
[tex]\displaystyle{=\frac{1}{1+1} }[/tex]
[tex]\displaystyle{=\frac{1}{2} }[/tex]
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KESIMPULAN
1. Hasil dari [tex]\displaystyle{ \lim_{t \to 2} \frac{(t^2-5t+6)sin(t-2)}{(t^2-t-2)^2}}[/tex] adalah [tex]\displaystyle{\boldsymbol{-\frac{1}{9} }}[/tex].
2. Hasil dari [tex]\displaystyle{\lim_{x \to -2} \frac{1-cos(x+2)}{x^2+4x+4}}[/tex] adalah [tex]\displaystyle{\boldsymbol{\frac{1}{2} }}[/tex].
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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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