Diketahui x→alimf(x)=−1 dan x→alimg(x)=6. Tentukan nilai: c. x→alim(g(x)2−6f(x)2f(x)⋅g(x))

Pertanyaan

Diketahui $\lim_{x\rightarrow a}f(x)=-1$ dan $\lim_{x\rightarrow a}g(x)=6$ .

Tentukan nilai:

c. $limit as x rightwards arrow a of open parentheses fraction numerator f left parenthesis x right parenthesis times g left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis squared minus 6 f left parenthesis x right parenthesis squared end fraction close parentheses$

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

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Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah $\boldsymbol-\frac{\mathbf1}{\mathbf5}$

Ingat!

Jika diketahui limit fungsi aljabar $\lim_{x\rightarrow a}f(x)=L$ dan $\lim_{x\rightarrow a}g(x)=M$ , serta bilangan real $c$ , maka:

$\lim_{x\rightarrow a}\left[f(x)+g(x)\right]=\lim_{x\rightarrow a}f(x)+\lim_{x\rightarrow a}g(x)=L+M$
$\lim_{x\rightarrow a}\left[cf(x)\right]=c\cdot\lim_{x\rightarrow a}f(x)=c\cdot L$
$\lim_{x\rightarrow a}\left(f(x)\right)^n=\left(\lim_{x\rightarrow a}f(x)\right)^n=L^n$
$\lim_{x\rightarrow a}\left[f(x)\cdot g(x)\right]=\lim_{x\rightarrow a}f(x)\cdot\lim_{x\rightarrow a}g(x)=L\cdot M$ \
$limit as x rightwards arrow a of open square brackets fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction close square brackets equals fraction numerator limit as x rightwards arrow a of f left parenthesis x right parenthesis over denominator limit as x rightwards arrow a of g left parenthesis x right parenthesis end fraction equals L over M$

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow a of open parentheses fraction numerator f left parenthesis x right parenthesis times g left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis squared minus 6 f left parenthesis x right parenthesis squared end fraction close parentheses end cell row blank equals cell fraction numerator limit as x rightwards arrow a of f left parenthesis x right parenthesis times g left parenthesis x right parenthesis over denominator limit as x rightwards arrow a of g left parenthesis x right parenthesis squared minus 6 times f left parenthesis x right parenthesis squared end fraction end cell row blank equals cell fraction numerator limit as x rightwards arrow a of f left parenthesis x right parenthesis times limit as x rightwards arrow a of g left parenthesis x right parenthesis over denominator open parentheses limit as x rightwards arrow a of g left parenthesis x right parenthesis close parentheses squared minus 6 times open parentheses limit as x rightwards arrow a of f left parenthesis x right parenthesis close parentheses squared end fraction end cell row blank equals cell fraction numerator open parentheses negative 1 close parentheses times 6 over denominator 6 squared minus 6 times open parentheses negative 1 close parentheses squared end fraction end cell row blank equals cell fraction numerator negative 6 over denominator 36 minus 6 end fraction end cell row blank equals cell fraction numerator negative 6 over denominator 30 end fraction end cell row blank equals cell fraction numerator negative 1 over denominator 5 end fraction end cell end table$

Dengan demikian, nilai limit dari $limit as x rightwards arrow a of open parentheses fraction numerator f left parenthesis x right parenthesis times g left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis squared minus 6 f left parenthesis x right parenthesis squared end fraction close parentheses$ , dengan $\lim_{x\rightarrow a}f(x)=-1$ dan $\lim_{x\rightarrow a}g(x)=6$ adalah $-\frac15$ .

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Idelia Arsanti

Makasih ❤️ Bantu banget

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