Roboguru

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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