Pertanyaan

Nilai dari x → ∞ lim ( 25 x 2 + 18 x + 2 − 5 x − 1 ) = ...

Nilai dari $x \to \infty lim (25 x^{2} + 18 x + 2 - 5 x - 1)$ = ...

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Bentuk limit fungsi: Jika nilai x disubstitusikan menghasilkan bentuk tak tentu maka perhitungannya dilakukan dengan mengalikan dengan faktor sekawannya. Tip! Jika a = p maka penyelesaiannya: Jadi,

Bentuk limit fungsi:

Jika nilai x disubstitusikan menghasilkan bentuk tak tentu maka perhitungannya dilakukan dengan mengalikan dengan faktor sekawannya.

$begin mathsize 12px style limit as x rightwards arrow infinity of open parentheses square root of 25 x squared plus 18 x plus 2 end root minus 5 x minus 1 close parentheses equals limit as x rightwards arrow infinity of open parentheses square root of 25 x squared plus 18 x plus 2 end root minus open parentheses 5 x minus 1 close parentheses close parentheses equals limit as x rightwards arrow infinity of open parentheses square root of 25 x squared plus 18 x plus 2 end root minus square root of open parentheses 5 x minus 1 close parentheses squared end root close parentheses equals limit as x rightwards arrow infinity of open parentheses square root of 25 x squared plus 18 x plus 2 end root minus square root of 25 x squared plus 10 x minus 1 end root close parentheses space space space space space times space open parentheses fraction numerator square root of 25 x squared plus 18 x plus 2 end root plus square root of 25 x squared plus 10 x plus 1 end root over denominator square root of 25 x squared plus 18 x plus 2 end root plus square root of 25 x squared plus 10 x plus 1 end root end fraction close parentheses equals limit as x rightwards arrow infinity of fraction numerator 25 x squared plus 18 x plus 2 minus open parentheses 25 x squared plus 10 x plus 1 close parentheses over denominator open parentheses square root of 25 x squared plus 18 x plus 2 end root plus square root of 25 x squared plus 10 x plus 1 end root close parentheses end fraction equals limit as x rightwards arrow infinity of fraction numerator 8 x plus 1 over denominator open parentheses square root of 25 x squared plus 18 x plus 2 end root plus square root of 25 x squared plus 10 x plus 1 end root close parentheses end fraction equals limit as x rightwards arrow infinity of fraction numerator 8 plus begin display style 1 over x end style over denominator open parentheses square root of 25 plus begin display style 18 over x end style plus begin display style 2 over x squared end style end root plus square root of 25 plus begin display style 10 over x end style plus begin display style 1 over x squared end style end root close parentheses end fraction equals fraction numerator 8 plus 0 over denominator square root of 25 plus 0 plus 0 end root plus square root of 25 plus 0 plus 0 end root end fraction equals fraction numerator 8 over denominator 5 plus 5 end fraction equals 8 over 10 equals 4 over 5 end style$

Tip!

Jika a = p maka penyelesaiannya: $begin mathsize 12px style fraction numerator b minus q over denominator 2 square root of a end fraction end style$

Jadi,

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$\style{font-size:14px}{\mathrm{Nilai}}\style{font-size:14px}\;\underset{\style{font-size:14px}{\mathrm x}\style{font-size:14px}\rightarrow\style{font-size:14px}\infty}{\style{font-size:14px}\lim}\left(\sqrt{\style{font-size:14px}4\style{font-size:14px}{\mathrm x}^\style{font-size:14px}2\style{font-size:14px}\;\style{font-size:14px}+\style{font-size:14px}\;\style{font-size:14px}3\style{font-size:14px}{\mathrm x}\style{font-size:14px}\;\style{font-size:14px}+\style{font-size:14px}\;\style{font-size:14px}4}\style{font-size:14px}\;\style{font-size:14px}-\style{font-size:14px}\;\style{font-size:14px}2\style{font-size:14px}{\mathrm x}\style{font-size:14px}\;\style{font-size:14px}+\style{font-size:14px}1\right)\style{font-size:14px}\;\style{font-size:14px}=\style{font-size:14px}\;\style{font-size:14px}.\style{font-size:14px}.\style{font-size:14px}.\style{font-size:14px}.$

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