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Pertanyaan

begin mathsize 14px style limit as x rightwards arrow negative 5 of space fraction numerator 2 x squared plus 7 x minus 15 over denominator 3 x squared plus 16 x plus 5 end fraction equals end style... 

  1. ...space 

  2. ...undefined 

Pembahasan Video:

Pembahasan Soal:

Gunakan pemfaktoran untuk mencari nilai dari limit seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 5 of space fraction numerator 2 x squared plus 7 x minus 15 over denominator 3 x squared plus 16 x plus 5 end fraction end cell equals cell limit as x rightwards arrow negative 5 of space fraction numerator 2 x squared plus 10 x minus 3 x minus 15 over denominator 3 x squared plus 15 x plus x plus 5 end fraction end cell row blank equals cell limit as x rightwards arrow negative 5 of fraction numerator 2 x left parenthesis x plus 5 right parenthesis minus 3 open parentheses x plus 5 close parentheses over denominator 3 x left parenthesis x plus 5 right parenthesis plus 1 left parenthesis x plus 5 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow negative 5 of fraction numerator left parenthesis 2 x minus 3 right parenthesis times down diagonal strike left parenthesis x plus 5 right parenthesis end strike over denominator left parenthesis 3 x plus 1 right parenthesis times down diagonal strike left parenthesis x plus 5 right parenthesis end strike end fraction end cell row blank equals cell limit as x rightwards arrow negative 5 of fraction numerator 2 x minus 3 over denominator 3 x plus 1 end fraction end cell end table end style

Kemudian substitusi undefined dengan begin mathsize 14px style negative 5 end style pada limit di atas.

begin mathsize 14px style limit as x rightwards arrow negative 5 of fraction numerator 2 x minus 3 over denominator 3 x plus 1 end fraction equals fraction numerator 2 open parentheses negative 5 close parentheses minus 3 over denominator 3 open parentheses negative 5 close parentheses plus 1 end fraction 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 10 minus 3 over denominator negative 15 plus 1 end fraction 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 13 over denominator negative 14 end fraction space space space space space space space space space space space space space space space space space space space space space equals 13 over 14 end style

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Freelancer6

Terakhir diupdate 03 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 5 of fraction numerator 2 x squared minus 17 x plus 35 over denominator x minus 5 end fraction end cell row blank equals cell limit as x rightwards arrow 5 of fraction numerator 2 x squared minus 17 x plus 35 over denominator x minus 5 end fraction end cell row blank equals cell limit as x rightwards arrow 5 of space fraction numerator left parenthesis 2 x minus 7 right parenthesis left parenthesis x minus 5 right parenthesis over denominator x minus 5 end fraction end cell row blank equals cell limit as x rightwards arrow 5 of space left parenthesis 2 x minus 7 right parenthesis end cell row blank equals cell 2 left parenthesis 5 right parenthesis minus 7 end cell row blank equals cell 10 minus 7 end cell row blank equals 3 end table end style 

Jadi, nilai dari begin mathsize 14px style limit as x rightwards arrow 5 of fraction numerator 2 x squared minus 17 x plus 35 over denominator x minus 5 end fraction end style adalah 3.

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Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 3 of fraction numerator x squared plus 4 x minus 21 over denominator x minus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator left parenthesis x minus 3 right parenthesis left parenthesis x plus 7 right parenthesis over denominator x minus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space x plus 7 end cell row blank equals cell 3 plus 7 end cell row blank equals 10 row blank blank blank end table end style 

Jadi, nilai dari begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator x squared plus 4 x minus 21 over denominator x minus 3 end fraction end style adalah 10.

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Roboguru

Pembahasan Soal:

Limit di atas akan diselesaikan dengan metode pemfaktoran terlebih dahulu karena saat diselesaikan menggunakan metode subtitusi secara langsung ditemukan hasil berbentuk tak tentu 0 over 0.

Dengan memfaktorkan fungsi dalam limit, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 5 of fraction numerator 2 x squared minus 17 x plus 35 over denominator x minus 5 end fraction end cell equals cell limit as x rightwards arrow 5 of fraction numerator open parentheses 2 x minus 7 close parentheses open parentheses x minus 5 close parentheses over denominator x minus 5 end fraction end cell row blank equals cell limit as x rightwards arrow 5 of open parentheses 2 x minus 7 close parentheses end cell row blank equals cell open parentheses 2 times 5 minus 7 close parentheses end cell row blank equals cell 10 minus 7 end cell row blank equals 3 end table

Jadi, limit as x rightwards arrow 5 of fraction numerator 2 x squared minus 17 x plus 35 over denominator x minus 5 end fraction equals 3.space 

0

Roboguru

Pembahasan Soal:

Limit di atas akan diselesaikan dengan metode pemfaktoran terlebih dahulu karena saat diselesaikan menggunakan metode subtitusi secara langsung ditemukan hasil berbentuk tak tentu 0 over 0.

Dengan memfaktorkan fungsi dalam limit, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator x squared plus 4 x minus 21 over denominator x minus 3 end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses x plus 7 close parentheses open parentheses x minus 3 close parentheses over denominator x minus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses x plus 7 close parentheses end cell row blank equals cell open parentheses 3 plus 7 close parentheses end cell row blank equals 10 end table 

Jadi, limit as x rightwards arrow 3 of fraction numerator x squared plus 4 x minus 21 over denominator x minus 3 end fraction equals 10.space 

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Roboguru

Nilai dari  adalah ....

Pembahasan Soal:

Permasalahan limit pada soal akan diselesaikan dengan metode pemfaktoran karena saat diselesaikan dengan menggunakan metode subtitusi secara langsung diperoleh hasil berbentuk tak tentu 0 over 0.

Dengan memfaktorkan fungsi dalam limit, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator x squared minus 1 over denominator x minus 1 end fraction end cell equals cell limit as x rightwards arrow 1 of fraction numerator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses over denominator x minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 1 of open parentheses x plus 1 close parentheses times stack space lim with x rightwards arrow 1 below open parentheses fraction numerator x minus 1 over denominator x minus 1 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 1 of open parentheses x plus 1 close parentheses times stack space lim with x rightwards arrow 1 below open parentheses 1 close parentheses end cell row blank equals cell limit as x rightwards arrow 1 of open parentheses x plus 1 close parentheses times 1 end cell row blank equals cell limit as x rightwards arrow 1 of open parentheses x plus 1 close parentheses end cell row blank equals cell 1 plus 1 end cell row blank equals 2 end table

Jadi, nilai dari limit as x rightwards arrow 1 of fraction numerator x squared minus 1 over denominator x minus 1 end fraction adalah 2.

Oleh karena itu, jawaban yang benar adalah E.space 

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