Nilai dari [tex]\displaystyle{ \lim_{x \to 1} \frac{f(x)}{g(x)+x^2-1} }[/tex] adalah D. -1.
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]
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DIKETAHUI
[tex]\displaystyle{ \lim_{x \to 1} \frac{f(x)}{g(x)}=3 }[/tex]
[tex]\displaystyle{ \lim_{x \to 1} \frac{x-1}{g(x)}=-2 }[/tex]
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DITANYA
Tentukan nilai dari [tex]\displaystyle{ \lim_{x \to 1} \frac{f(x)}{g(x)+x^2-1} }[/tex]
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PENYELESAIAN
[tex]\displaystyle{ \lim_{x \to 1} \frac{f(x)}{g(x)}=3 }[/tex]
[tex]\displaystyle{\frac{\lim\limits_{x \to 1} f(x)}{\lim\limits_{x \to 1} g(x)}=3 }[/tex]
[tex]\displaystyle{\lim\limits_{x \to 1} f(x)=3\lim\limits_{x \to 1} g(x)~~~...(i) }[/tex]
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[tex]\displaystyle{ \lim_{x \to 1} \frac{x-1}{g(x)}=-2 }[/tex]
[tex]\displaystyle{\frac{\lim\limits_{x \to 1} (x-1)}{\lim\limits_{x \to 1} g(x)}=-2 }[/tex]
[tex]\displaystyle{\frac{1-1}{g(1)}=-2 }[/tex]
[tex]\displaystyle{\frac{0}{g(1)}\neq -2 }[/tex]
Karena hasil dari substitusi langsung nilainya tidak konsisten, maka dapat kita simpulkan bahwa fungsi g(x) merupakan fungsi yang mempunyai faktor (x-1), bisa kita tulis : [tex]g(x)=H(x)(x-1)~~~...(ii)[/tex]
[tex]\displaystyle{ \lim_{x \to 1} \frac{x-1}{g(x)}=-2 }[/tex]
[tex]\displaystyle{ \lim_{x \to 1} \frac{\cancel{(x-1)}}{H(x)\cancel{(x-1)}}=-2 }[/tex]
[tex]\displaystyle{ \lim_{x \to 1} \frac{1}{H(x)}=-2 }[/tex]
[tex]\displaystyle{\frac{1}{\lim\limits_{x \to 1} H(x)}=-2 }[/tex]
[tex]\displaystyle{\lim\limits_{x \to 1} H(x)=-\frac{1}{2}~~~...(iii) }[/tex]
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Sehingga :
[tex]\displaystyle{ \lim_{x \to 1} \frac{f(x)}{g(x)+x^2-1} }[/tex]
[tex]\displaystyle{=\lim_{x \to 1} \frac{f(x)}{H(x)(x-1)+(x+1)(x-1)} }[/tex]
[tex]\displaystyle{=\lim_{x \to 1} \frac{f(x)}{(x-1)[H(x)+(x+1)]} }[/tex]
[tex]\displaystyle{=\frac{\lim\limits_{x \to 1} f(x)}{\lim\limits_{x \to 1} (x-1)[H(x)+(x+1)]} }[/tex]
[tex]\displaystyle{=\frac{3\lim\limits_{x \to 1} g(x)}{\lim\limits_{x \to 1} (x-1)[H(x)+(x+1)]} }[/tex]
[tex]\displaystyle{=\frac{3\lim\limits_{x \to 1} H(x)\cancel{(x-1)}}{\lim\limits_{x \to 1} \cancel{(x-1)}[H(x)+(x+1)]} }[/tex]
[tex]\displaystyle{=\frac{3\lim\limits_{x \to 1} H(x)}{\lim\limits_{x \to 1} [H(x)+(x+1)]} }[/tex]
[tex]\displaystyle{=\frac{3\lim\limits_{x \to 1} H(x)}{\lim\limits_{x \to 1} H(x)+\lim\limits_{x \to 1} (x+1)} }[/tex]
[tex]\displaystyle{=\frac{3\times\left ( -\frac{1}{2} \right )}{-\frac{1}{2}+1+1} }[/tex]
[tex]\displaystyle{=\frac{-\frac{3}{2}}{\frac{3}{2}} }[/tex]
[tex]=-1[/tex]
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KESIMPULAN
Nilai dari [tex]\displaystyle{ \lim_{x \to 1} \frac{f(x)}{g(x)+x^2-1} }[/tex] adalah D. -1.
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PELAJARI LEBIH LANJUT
- Limit fungsi : https://brainly.co.id/tugas/29558741
- Limit tak hingga : https://brainly.co.id/tugas/38915286
- Limit trigonometri : https://brainly.co.id/tugas/38915095
- Limit teorema apit : https://brainly.co.id/tugas/35849860
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DETAIL JAWABAN
Kelas : 11
Mapel: Matematika
Bab : Limit Fungsi
Kode Kategorisasi: 11.2.8
Kata Kunci : limit, fungsi.
Jawab:
limit
[tex]\sf (i). ~lim_{x\to 1}~ \dfrac{f(x)}{g(x)} = 3 , maka f(x) = 3.g(x)[/tex]
[tex]\sf (ii). ~lim_{x\to 1}~ \dfrac{x- 1}{g(x)} = -2[/tex]
x= 1 , bentuk 0/0
[tex]\sf maka ~lim_{x\to 1}~ \dfrac{(x- 1)(x- 3)}{(x-1)} = -2\to g(x) = \dfrac{x-1}{x-3}[/tex]
[tex]\sf f(x)= 3 g(x) = \dfrac{3(x - 1)}{x-3}[/tex]
hasil dari
[tex]\sf ~lim_{x\to 1}~ \dfrac{f(x)}{g(x)+ x^2 - 1} = . .[/tex]
[tex]\sf lim_{x\to 1}~ \dfrac{\frac{3(x-1)}{x-3}}{\frac{x-1}{x-3} + (x-1)(x+1}[/tex]
*kalikan (x- 3)/(x-3)
[tex]\sf lim_{x\to 1}~ \dfrac{3(x -1)}{(x-1)+ (x -1)(x+ 1)(x -3)}[/tex]
[tex]\sf lim_{x\to 1}~ \dfrac{3(x -1)}{(x-1)\{1 + (x+ 1)(x -3)\}}[/tex]
[tex]\sf lim_{x\to 1}~ \dfrac{3}{\{1 + (x+ 1)(x -3)\}}[/tex]
x= 1
[tex]\sf lim_{x\to 1}~ \dfrac{3}{\{1 + (1+ 1)(1 -3)\}}[/tex]
[tex]\sf =\dfrac{3}{1 + 2(-2)} = \dfrac{3}{-3}= -1[/tex]
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