Roboguru

Pertanyaan

begin mathsize 14px style limit as x rightwards arrow 9 of fraction numerator x minus 9 over denominator square root of x minus 3 end fraction equals... end style 

Pembahasan Video:

Pembahasan Soal:

Perhatikan perhitungan berikut ini!

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 9 of fraction numerator x minus 9 over denominator square root of x minus 3 end fraction end cell equals cell limit as x rightwards arrow 9 of fraction numerator open parentheses square root of x plus 3 close parentheses open parentheses square root of x minus 3 close parentheses over denominator square root of x minus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 9 of square root of x plus 3 end cell row blank equals cell square root of 9 plus 3 end cell row blank equals cell 3 plus 3 end cell row blank equals 6 end table end style

Dengan demikian, begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 9 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x minus 9 over denominator square root of x minus 3 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 6 end table end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Freelancer9

Terakhir diupdate 03 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai

Pembahasan Soal:

Substitusi nilai x equals 1, diperoleh 

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator x to the power of 4 minus x end fraction end cell equals cell space fraction numerator 1 minus square root of 1 over denominator open parentheses 1 close parentheses to the power of 4 minus open parentheses 1 close parentheses end fraction end cell row blank equals cell 0 over 0 end cell end table 

Hasilnya bentuk tak tentu, maka limit tersebut dapat diselesaikan dengan cara pemfaktoran seperti di bawah ini 

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator x to the power of 4 minus x end fraction end cell equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator x open parentheses x cubed minus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator x open parentheses x cubed minus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator x open parentheses x minus 1 close parentheses open parentheses x squared plus x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator x open parentheses square root of x plus 1 close parentheses open parentheses square root of x minus 1 close parentheses open parentheses x squared plus x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator negative x open parentheses square root of x plus 1 close parentheses open parentheses 1 minus square root of x close parentheses open parentheses x squared plus x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of minus fraction numerator 1 over denominator x open parentheses square root of x plus 1 close parentheses open parentheses x squared plus x plus 1 close parentheses end fraction end cell row blank equals cell negative fraction numerator 1 over denominator open parentheses 1 close parentheses open parentheses square root of 1 plus 1 close parentheses open parentheses open parentheses 1 close parentheses squared plus open parentheses 1 close parentheses plus 1 close parentheses end fraction end cell row blank equals cell negative fraction numerator 1 over denominator open parentheses 1 close parentheses open parentheses 2 close parentheses open parentheses 3 close parentheses end fraction end cell row blank equals cell negative 1 over 6 end cell end table 

Dengan demikian, nilai limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator x to the power of 4 minus x end fraction equals negative 1 over 6

Roboguru

Nilai dari  adalah ...

Pembahasan Soal:

Dengan melakukan pemfaktoran terlebih dahulu diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator x squared minus x minus 6 over denominator x minus 3 end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses x minus 3 close parentheses open parentheses x plus 2 close parentheses over denominator open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses x plus 2 close parentheses end cell row blank equals cell open parentheses 3 close parentheses plus 2 end cell row blank equals 5 end table end style 

Jadi begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator x squared minus x minus 6 over denominator x minus 3 end fraction equals 5 end style

Roboguru

Nilai

Pembahasan Soal:

Penyelesaian limit di atas dapat digunakan metode pemfaktoran.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator 2 x squared plus 2 x minus 12 over denominator x squared minus 6 x plus 8 end fraction end cell equals cell limit as x rightwards arrow 2 of fraction numerator up diagonal strike open parentheses x minus 2 close parentheses end strike open parentheses 2 x plus 6 close parentheses over denominator open parentheses x minus 4 close parentheses up diagonal strike open parentheses x minus 2 close parentheses end strike end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator left parenthesis 2 x plus 6 right parenthesis over denominator left parenthesis x minus 4 right parenthesis end fraction end cell row blank equals cell fraction numerator 2 open parentheses 2 close parentheses plus 6 over denominator 2 minus 4 end fraction end cell row blank equals cell fraction numerator 10 over denominator negative 2 end fraction end cell row blank equals cell negative 5 end cell end table end style 

Dengan demikian diperoleh begin mathsize 14px style limit as x rightwards arrow 2 of fraction numerator 2 x squared plus 2 x minus 12 over denominator x squared minus 6 x plus 8 end fraction equals negative 5 end style.

Jadi, jawaban yang tepat adalah A.

Roboguru

Nilai  adalah ...

Pembahasan Soal:

begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator x to the power of 5 minus 3 x to the power of 4 over denominator x to the power of 6 minus 7 x to the power of 5 plus 10 x to the power of 4 end fraction equals limit as x rightwards arrow 0 of fraction numerator x to the power of 4 open parentheses x minus 3 close parentheses over denominator x to the power of 4 open parentheses x squared minus 7 x squared plus 10 close parentheses end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 0 minus 3 over denominator 0 minus 0 plus 10 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator negative 3 over denominator 10 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals negative 0 comma 3 end style  

jadi, jawaban yang yang tepat D

Roboguru

Nilai adalah ...

Pembahasan Soal:

begin mathsize 14px style limit as x rightwards arrow 4 of fraction numerator 4 x squared minus 64 over denominator x plus square root of x minus 6 end fraction equals limit as x rightwards arrow 4 of fraction numerator 4 open parentheses x plus 4 close parentheses open parentheses x minus 4 close parentheses over denominator open parentheses square root of x plus 3 close parentheses open parentheses square root of x minus 2 close parentheses end fraction space space space space space space space space space space space space space space space space space space space space space space space equals limit as x rightwards arrow 4 of fraction numerator 4 open parentheses x plus 4 close parentheses open parentheses square root of x plus 2 close parentheses open parentheses square root of x minus 2 close parentheses over denominator open parentheses square root of x plus 3 close parentheses open parentheses square root of x minus 2 close parentheses end fraction space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 4 open parentheses 4 plus 4 close parentheses open parentheses square root of 4 plus 2 close parentheses over denominator square root of 4 plus 3 end fraction space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 4 open parentheses 8 close parentheses open parentheses 4 close parentheses over denominator 2 plus 3 end fraction space space space space space space space space space space space space space space space space space space space space space space space equals 128 over 5 end style 

Jadi, jawaban yang tepat adalah A

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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