Pertanyaan

x → 2 lim 3 − x 2 + 5 ( 4 − x 2 ) = ...

$x \to 2 lim \frac{( 4 - x ^{2} )}{3 - x ^{2} + 5} = ...$

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Klaim

S. Indah

Master Teacher

Mahasiswa/Alumni Universitas Lampung

Jawaban terverifikasi

Jawaban

nilai dari adalah 6.

nilai dari $begin mathsize 14px style limit as x rightwards arrow 2 of space fraction numerator open parentheses 4 minus x squared close parentheses over denominator 3 minus square root of x squared plus 5 end root end fraction end style$ adalah 6.

Pembahasan

Dengan menerapkanpenyelesaian limit dengan metode mengalikan akar sekawan, diperoleh perhitungan sebagai berikut. Dengan demikian, nilai dari adalah 6.

Dengan menerapkan penyelesaian limit dengan metode mengalikan akar sekawan, diperoleh perhitungan sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 2 of space fraction numerator open parentheses 4 minus x squared close parentheses over denominator 3 minus square root of x squared plus 5 end root end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses 4 minus x squared close parentheses over denominator 3 minus square root of x squared plus 5 end root end fraction cross times fraction numerator 3 plus square root of x squared plus 5 end root over denominator 3 plus square root of x squared plus 5 end root end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses 4 minus x squared close parentheses open parentheses 3 plus square root of x squared plus 5 end root close parentheses over denominator 3 squared minus open parentheses square root of x squared plus 5 end root close parentheses squared end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses 4 minus x squared close parentheses open parentheses 3 plus square root of x squared plus 5 end root close parentheses over denominator 9 minus open parentheses x squared plus 5 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator up diagonal strike open parentheses 4 minus x squared close parentheses end strike open parentheses 3 plus square root of x squared plus 5 end root close parentheses over denominator up diagonal strike open parentheses 4 minus x squared close parentheses end strike end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space open parentheses 3 plus square root of x squared plus 5 end root close parentheses end cell row blank equals cell 3 plus square root of 2 squared plus 5 end root end cell row blank equals cell 3 plus square root of 9 end cell row blank equals cell 3 plus 3 end cell row blank equals 6 end table end style$

Dengan demikian, nilai dari $begin mathsize 14px style limit as x rightwards arrow 2 of space fraction numerator open parentheses 4 minus x squared close parentheses over denominator 3 minus square root of x squared plus 5 end root end fraction end style$ adalah 6.