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Monday, April 1, 2019

Teorema Limit Matematika




Dengan $n$ adalah bilangan bulat positif, $k$ konstanta, $f$ dan $g$ adalah fungsi yang mempunyai limit di $c$. Maka :

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Teorema 1:
$\boxed{\displaystyle\lim_{x \to c}k=k}$

Contoh:
$\displaystyle\lim_{x \to c}5=5$

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Teorema 2:
$\boxed{\displaystyle\lim_{x \to c}x=c}$

Contoh:
$\displaystyle\lim_{x \to 5}x=5$

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Teorema 3:
$\boxed{\displaystyle\lim_{x \to c}kf(x)=k \lim_{x \to c}f(x)}$

Contoh:
$\displaystyle\lim_{x \to c}2x^2$
$=\displaystyle {2\lim_{x \to 5}x^2}$
$=\displaystyle {2.(5^2)}$
$=\displaystyle {20}$

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Teorema 4:
$\boxed{\displaystyle\lim_{x \to c}[f(x)+g(x)]=\lim_{x \to c}f(x)+ \lim_{x \to c}g(x)}$

Contoh:
$\displaystyle\lim_{x \to 5}2x^2+7x$
$=\displaystyle{2\lim_{x \to 5}x^2+7\lim_{x \to 5}x}$
$=\displaystyle{2.5^2+7.5}$
$=\displaystyle{55}$

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Teorema 5:
$\boxed{\displaystyle\lim_{x \to c}[f(x)-g(x)]=\lim_{x \to c}f(x)- \lim_{x \to c}g(x)}$

Contoh:
$\displaystyle\lim_{x \to 5}x^3+4x$
$=\displaystyle{\lim_{x \to 2}x^3+4\lim_{x \to 2}x}$
$=\displaystyle{2^3+4.2}$
$=\displaystyle{16}$

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Teorema 6:
$\boxed{\displaystyle\lim_{x \to c}[f(x).g(x)]=(\lim_{x \to c}f(x)).( \lim_{x \to c}g(x))}$

Contoh:
$\displaystyle\lim_{x \to 5}(x-1)(2x-3)$
$=\displaystyle{\lim_{x \to 5}(x-1).\lim_{x \to 5}(2x-3)}$
$=\displaystyle{(5-1)((2.5)-3}$
$=\displaystyle{4.7}$
$=\displaystyle{28}$

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Teorema 7:
$\boxed{\displaystyle{{\displaystyle\lim_{x \to c}\frac{f(x)}{g(x)}=\frac{\displaystyle\lim_{x \to c}f(x)}{\displaystyle\lim_{x \to c}g(x)}}}}$

Dengan $\displaystyle\lim_{x \to c}g(x)\neq 0$

Contoh:
$\displaystyle\lim_{x \to 5}\frac{(x^2-5)}{(2x-1)}$
$=\displaystyle{\frac{\lim_{x \to 5}(x^2-5)}{\lim_{x \to 5}(2x-1)}}$
$=\displaystyle{\frac{(5^2-5)}{((2.5)-1)}}$
$=\displaystyle{\frac{20}{9}}$

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Teorema 8:
$\boxed{\displaystyle\lim_{x \to c}(f(x))^2=(\lim_{x \to c}f(x))^2}$

Contoh:
$\displaystyle\lim_{x \to 2}(x^2-1)^3$
$=\displaystyle{(\lim_{x \to 2}(x^2-1))^3}$
$=\displaystyle{(2^2-1)^3}$
$=\displaystyle{3^3}$
$=\displaystyle{27}$

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Teorema 9:
$\boxed{\displaystyle\lim_{x \to c}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x \to c}f(x)}}$

Dengan $\displaystyle\lim_{x \to c}f(x)>0$ jika n bilangan genap

Contoh:
$\displaystyle\lim_{x \to 3}\sqrt[2]{(x^3-2)}$
$=\displaystyle\sqrt[2]{\lim_{x \to 3}(x^3-2)}$
$=\displaystyle{\sqrt[2]{(3^3-2)}}$
$=\displaystyle{\sqrt[2]{25}}$
$=\displaystyle{5}$

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Sekian dan terima kasih..semoga bermanfaat...

Referensi: Kalkulus edisi Ketujuh oleh Dale Varberg dan Edwin J. Purcell

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