Nilai limit dari adalah ....

Pertanyaan

Nilai limit dari $limit as n rightwards arrow 1 of space fraction numerator 1 minus x over denominator 2 minus square root of x plus 3 end root end fraction$ adalah ....

Pembahasan Soal:

Asumsikan bahwa $n$ pada soal merupakan $x$ .

Perhatikan perhitungan berikut!

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of space fraction numerator 1 minus x over denominator 2 minus square root of x plus 3 end root end fraction end cell equals cell fraction numerator 1 minus 1 over denominator 2 minus square root of 1 plus 3 end root end fraction end cell row blank equals cell fraction numerator 1 minus 1 over denominator 2 minus 2 end fraction end cell row blank equals cell 0 over 0 end cell end table$

Karena hasil dari limit tersebut ketika disubstitusikan $x equals 1$ merupakan bilangan tak tentu $0 over 0$ , maka kita kalikan bentuk limit tersebut dengan akar sekawan.
Dengan demikian, diperoleh hasil perhitungan sebagai berikut.

$table attributes columnalign right center left columnspacing 2px end attributes row cell limit as x rightwards arrow 1 of space fraction numerator 1 minus x over denominator 2 minus square root of x plus 3 end root end fraction end cell equals cell limit as x rightwards arrow 1 of space open parentheses fraction numerator 1 minus x over denominator 2 minus square root of x plus 3 end root end fraction cross times fraction numerator 2 plus square root of x plus 3 end root over denominator 2 plus square root of x plus 3 end root end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 1 of space open parentheses fraction numerator open parentheses 1 minus x close parentheses open parentheses 2 plus square root of x plus 3 end root close parentheses over denominator 4 minus open parentheses x plus 3 close parentheses end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 1 of space open parentheses fraction numerator up diagonal strike open parentheses 1 minus x close parentheses end strike open parentheses 2 plus square root of x plus 3 end root close parentheses over denominator up diagonal strike open parentheses 1 minus x close parentheses end strike end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 1 of space open parentheses 2 plus square root of x plus 3 end root close parentheses end cell row blank equals cell 2 plus square root of 1 plus 3 end root end cell row blank equals cell 2 plus 2 end cell row blank equals 4 end table$

Jadi, diperoleh bahwa $limit as x rightwards arrow 1 of fraction numerator 1 minus x over denominator 2 minus square root of x plus 3 end root end fraction equals 4$ .

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 18 Juli 2021

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