Pertanyaan

Nilai dari x → ∞ lim ( ( 2 x + 1 ) ( 2 x − 5 ) − ( 2 x − 5 ) ) adalah ....

Nilai dari $x \to \infty lim ((2 x + 1) (2 x - 5) - (2 x - 5))$ adalah ....

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Jadi, jawaban yang tepat adalah E.

$begin mathsize 11px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of left parenthesis square root of left parenthesis 2 x plus 1 right parenthesis left parenthesis 2 x minus 5 right parenthesis end root minus left parenthesis 2 x minus 5 right parenthesis right parenthesis space end cell row blank equals cell limit as x rightwards arrow infinity of left parenthesis square root of 4 x squared minus 8 x minus 5 end root minus left parenthesis 2 x minus 5 right parenthesis right parenthesis space end cell row blank equals cell limit as x rightwards arrow infinity of left parenthesis square root of 4 x squared minus 8 x minus 5 end root minus left parenthesis 2 x minus 5 right parenthesis right parenthesis space cross times open parentheses fraction numerator square root of 4 x squared minus 8 x minus 5 end root plus left parenthesis 2 x minus 5 right parenthesis space over denominator square root of 4 x squared minus 8 x minus 5 end root plus left parenthesis 2 x minus 5 right parenthesis space end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 4 x squared minus 8 x minus 5 minus left parenthesis 2 x minus 5 right parenthesis squared over denominator square root of 4 x squared minus 8 x minus 5 end root plus left parenthesis 2 x minus 5 right parenthesis space end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 4 x squared minus 8 x minus 5 minus left parenthesis 4 x squared minus 20 x plus 25 right parenthesis over denominator square root of x squared left parenthesis 4 minus begin display style 8 over x end style minus begin display style 5 over x squared end style right parenthesis end root plus left parenthesis 2 x minus 5 right parenthesis space end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 12 x minus 30 over denominator x square root of 4 minus begin display style 8 over x end style minus begin display style 5 over x squared end style end root plus x left parenthesis 2 minus begin display style 5 over x end style right parenthesis space end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator x left parenthesis 12 minus 30 over x right parenthesis over denominator x open parentheses square root of 4 minus 8 over x minus 5 over x squared end root plus left parenthesis 2 minus 5 over x right parenthesis close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 12 minus 30 over x over denominator square root of 4 minus 8 over x minus 5 over x squared end root plus left parenthesis 2 minus 5 over x right parenthesis end fraction end cell row blank equals cell fraction numerator 12 minus 0 over denominator square root of 4 minus 0 minus 0 end root plus left parenthesis 2 minus 0 right parenthesis end fraction end cell row blank equals cell fraction numerator 12 over denominator 2 plus 2 end fraction end cell row blank equals 3 end table end style$

Jadi, jawaban yang tepat adalah E.