Roboguru

Nilai x→3lim​4−x2+7​9−x2​=....

Pertanyaan

Nilai limit as x rightwards arrow 3 of fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction equals....

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction end cell equals cell fraction numerator 9 minus open parentheses 3 close parentheses squared over denominator 4 minus square root of open parentheses 3 close parentheses squared plus 7 end root end fraction end cell row blank equals cell fraction numerator 9 minus 9 over denominator 4 minus square root of 16 end fraction end cell row blank equals cell 0 over 0 end cell end table
 

Berdasarkan uraian di atas, dengan metode substitusi langsung, diperoleh hasil limitnya adalah 0 over 0, maka nilai limit dapat ditentukan dengan metode mengalikan akar sekawan terlebih dahulu sebagai berikut.
 

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction end cell equals cell limit as x rightwards arrow 3 of open parentheses fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction cross times fraction numerator 4 plus square root of x squared plus 7 end root over denominator 4 plus square root of x squared plus 7 end root end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses fraction numerator 9 minus x squared left parenthesis 4 plus square root of x squared plus 7 end root right parenthesis over denominator 16 minus left parenthesis x squared plus 7 right parenthesis end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses fraction numerator up diagonal strike 9 minus x squared end strike left parenthesis 4 plus square root of x squared plus 7 end root right parenthesis over denominator up diagonal strike 9 minus x squared end strike end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses 4 plus square root of x squared plus 7 end root close parentheses end cell row blank equals cell 4 plus square root of open parentheses 3 close parentheses squared plus 7 end root end cell row blank equals cell 4 plus 4 end cell row blank equals 8 end table


Jadi, limit as x rightwards arrow 3 of fraction numerator 9 minus x squared over denominator 4 minus square root of x squared plus 7 end root end fraction equals 8.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Nur

Terakhir diupdate 07 Oktober 2021

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Pertanyaan yang serupa

Nilai dari x→−2lim​x+6​−2x+2​ adalah ....

Pembahasan Soal:

Jika disusubstitusikan x equals negative 2, didapat nilai dari limit as x rightwards arrow negative 2 of fraction numerator x plus 2 over denominator square root of x plus 6 end root minus 2 end fraction adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 of fraction numerator x plus 2 over denominator square root of x plus 6 end root minus 2 end fraction end cell equals cell fraction numerator limit as x rightwards arrow negative 2 of open parentheses x plus 2 close parentheses over denominator limit as x rightwards arrow negative 2 of open parentheses square root of x plus 6 end root minus 2 close parentheses end fraction end cell row blank equals cell fraction numerator negative 2 plus 2 over denominator square root of negative 2 plus 6 end root minus 2 end fraction end cell row blank equals cell fraction numerator negative 2 plus 2 over denominator square root of 4 minus 2 end fraction end cell row blank equals cell fraction numerator negative 2 plus 2 over denominator 2 minus 2 end fraction end cell row blank equals cell 0 over 0 end cell end table

Didapat bentuk undefined . Ingat bahwa undefined merupakan bentuk tak tentu. Oleh karena itu, perlu cara lain untuk mencari nilai limitnya, yaitu dengan mengalikan dengan faktor sekawan.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 of fraction numerator x plus 2 over denominator square root of x plus 6 end root minus 2 end fraction end cell equals cell limit as x rightwards arrow negative 2 of open parentheses fraction numerator x plus 2 over denominator square root of x plus 6 end root minus 2 end fraction times fraction numerator square root of x plus 6 end root plus 2 over denominator square root of x plus 6 end root plus 2 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow negative 2 of fraction numerator open parentheses x plus 2 close parentheses open parentheses square root of x plus 6 end root plus 2 close parentheses over denominator open parentheses square root of x plus 6 end root close parentheses squared minus 2 squared end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of fraction numerator open parentheses x plus 2 close parentheses open parentheses square root of x plus 6 end root plus 2 close parentheses over denominator open parentheses x plus 6 close parentheses minus 4 end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of fraction numerator open parentheses x plus 2 close parentheses open parentheses square root of x plus 6 end root plus 2 close parentheses over denominator x plus 2 end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of fraction numerator open parentheses x plus 2 close parentheses open parentheses square root of x plus 6 end root plus 2 close parentheses over denominator x plus 2 end fraction end cell row blank equals cell limit as x rightwards arrow negative 2 of open parentheses square root of x plus 6 end root plus 2 close parentheses end cell row blank equals cell square root of negative 2 plus 6 end root plus 2 end cell row blank equals cell square root of 4 plus 2 end cell row blank equals cell 2 plus 2 end cell row blank equals 4 end table

Dengan demikian, nilai dari begin mathsize 14px style limit as straight x rightwards arrow negative 2 of fraction numerator straight x plus 2 over denominator square root of straight x plus 6 end root minus 2 end fraction end style adalah 4.

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Nilai x→−2lim​x+25x2+9x−2​=...

Pembahasan Soal:

Dengan cara di faktorkan

limit as x rightwards arrow negative 2 of fraction numerator 5 x squared plus 9 x minus 2 over denominator x plus 2 end fraction  limit as x rightwards arrow negative 2 of fraction numerator begin display style left parenthesis 5 x minus 1 right parenthesis left parenthesis x plus 2 right parenthesis end style over denominator begin display style x plus 2 end style end fraction  limit as x rightwards arrow negative 2 of 5 x minus 1  equals 5 left parenthesis negative 2 right parenthesis minus 1  equals negative 11

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Roboguru

Nilai x→3lim​6=...

Pembahasan Soal:

Ingat sifat limit fungsi berikut.

xclimk=k 

Maka penyelesaian soal di atas yaitu

begin mathsize 14px style limit as x rightwards arrow 3 of 6 equals 6 end style 

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Nilai x→1lim​x−12x2−7x+5​=....

Pembahasan Soal:

Dengan cara difaktorkan

limit as straight x rightwards arrow 1 of fraction numerator 2 straight x squared minus 7 straight x plus 5 over denominator straight x minus 1 end fraction  limit as straight x rightwards arrow 1 of fraction numerator left parenthesis 2 straight x minus 5 right parenthesis left parenthesis straight x minus 1 over denominator left parenthesis straight x minus 1 right parenthesis end fraction  limit as straight x rightwards arrow 1 of 2 straight x minus 5  space space space space space space space equals 2 left parenthesis 1 right parenthesis minus 5  space space space space space space space equals negative 3

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Roboguru

x→−5lim​4x2+2x−1=...

Pembahasan Soal:

Perhatikan perhitungan berikut ini

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 5 of space 4 x squared plus 2 x minus 1 end cell equals cell 4 open parentheses negative 5 close parentheses squared plus 2 open parentheses negative 5 close parentheses minus 1 end cell row blank equals cell 4 open parentheses 25 close parentheses minus 10 minus 1 end cell row blank equals cell 100 minus 10 minus 1 end cell row blank equals 89 end table

Dengan demikian hasil dari limit as x rightwards arrow negative 5 of space 4 x squared plus 2 x minus 1 adalah 89.

0

Roboguru

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