Roboguru

Nilai

Pertanyaan

Nilai limit as x rightwards arrow negative 3 of space open parentheses fraction numerator 5 over denominator x plus 3 end fraction plus fraction numerator 30 over denominator x squared minus 9 end fraction close parentheses equals space horizontal ellipsis 

Pembahasan Soal:

Nilai limit tersebut dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow negative 3 of space open parentheses fraction numerator 5 over denominator x plus 3 end fraction plus fraction numerator 30 over denominator x squared minus 9 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow negative 3 of space open parentheses fraction numerator 5 over denominator x plus 3 end fraction plus fraction numerator 30 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow negative 3 of space open parentheses fraction numerator 5 open parentheses x minus 3 close parentheses over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction plus fraction numerator 30 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow negative 3 of space open parentheses fraction numerator 5 x plus 15 over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow negative 3 of space fraction numerator 5 open parentheses x plus 3 close parentheses over denominator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 3 of space fraction numerator 5 over denominator open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell fraction numerator 5 over denominator negative 3 minus 3 end fraction end cell row blank equals cell negative 5 over 6 end cell end table

Dengan demikian, nilai table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow negative 3 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses fraction numerator 5 over denominator x plus 3 end fraction plus fraction numerator 30 over denominator x squared minus 9 end fraction close parentheses 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 minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 5 over 6 end cell end table 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Eka

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 05 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

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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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Roboguru

= ...

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.

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 

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 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

Tentukan nilai

Pembahasan Soal:

begin mathsize 14px style limit as x rightwards arrow 2 of space fraction numerator 3 x squared minus 8 x plus 4 over denominator x squared minus 5 x plus 6 end fraction equals limit as x rightwards arrow 2 of space fraction numerator open parentheses 3 x minus 2 close parentheses open parentheses x minus 2 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses 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 space space space space space space equals limit as x rightwards arrow 2 of space fraction numerator 3 x minus 2 over denominator x minus 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 space space space space space space equals fraction numerator 3 open parentheses 2 close parentheses minus 2 over denominator 2 minus 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 space space space space space space equals fraction numerator 6 minus 2 over denominator 2 minus 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 space space space space space space equals fraction numerator 4 over denominator negative 1 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 equals negative 4 end style  

Jadi, begin mathsize 14px style limit as x rightwards arrow 2 of space fraction numerator 3 x squared minus 8 x plus 4 over denominator x squared minus 5 x plus 6 end fraction equals negative 4 end style.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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