Roboguru

Hitung nilai

Pertanyaan

Hitung nilai limit as x rightwards arrow 1 of space open parentheses fraction numerator 2 over denominator x squared minus 1 end fraction minus fraction numerator 1 over denominator x minus 1 end fraction close parentheses equals space horizontal ellipsis 

Pembahasan Soal:

Nilai limit fungsi tersebut dapat ditentukan sebagai berikut.

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

0

Roboguru

Nilai  = ....

Pembahasan Soal:

Gunakan perkalian sekawan.

begin mathsize 14px style limit with x rightwards arrow 3 below fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction end style = begin mathsize 14px style limit with x rightwards arrow 3 below fraction numerator 2 minus square root of x plus 1 end root over denominator x minus 3 end fraction cross times fraction numerator 2 plus square root of x plus 1 end root over denominator 2 plus square root of x plus 1 end root end fraction end style  

begin mathsize 14px style limit with x rightwards arrow 3 below fraction numerator 4 minus left parenthesis x plus 1 right parenthesis over denominator left parenthesis x minus 3 right parenthesis left parenthesis 2 plus square root of x plus 1 end root right parenthesis space end fraction end style 

begin mathsize 14px style limit with x rightwards arrow 3 below fraction numerator negative open parentheses x minus 3 close parentheses over denominator left parenthesis x minus 3 right parenthesis left parenthesis 2 plus square root of x plus 1 end root right parenthesis space end fraction end style 

begin mathsize 14px style fraction numerator negative 1 over denominator 2 plus square root of 4 end fraction equals fraction numerator negative 1 over denominator 4 end fraction end style.

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Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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