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Pertanyaan

begin mathsize 14px style limit as x rightwards arrow 3 of left parenthesis 3 x minus 5 right parenthesis cubed equals... end style 

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of left parenthesis 3 x minus 5 right parenthesis cubed end cell equals cell open parentheses 3 left parenthesis 3 right parenthesis minus 5 close parentheses cubed end cell row blank equals cell 4 cubed end cell row blank equals 64 end table end style

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

Nilai  adalah...

Pembahasan Soal:

Dengan menggunakan konsep dasar limit diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 5 of left parenthesis 3 x squared minus 15 x plus 10 right parenthesis end cell equals cell 3 open parentheses 5 close parentheses squared minus 15 open parentheses 5 close parentheses plus 10 end cell row blank equals cell 3 open parentheses 25 close parentheses minus 75 plus 10 end cell row blank equals cell 75 minus 75 plus 10 end cell row blank equals 10 end table end style

Dengan demikian nilai dari begin mathsize 14px style limit as x rightwards arrow 5 of left parenthesis 3 x squared minus 15 x plus 10 right parenthesis end style adalah begin mathsize 14px style 10 end style.

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Roboguru

Nilai  adalah

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of 2 x squared plus x minus 3 end cell equals cell 2 left parenthesis 2 right parenthesis squared plus 2 minus 3 end cell row blank equals cell 2 open parentheses 4 close parentheses plus 2 minus 3 end cell row blank equals cell 10 minus 3 end cell row blank equals 7 end table

Jadi, nilai limit as x rightwards arrow 2 of 2 x squared plus x minus 3 adalah 7.

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Roboguru

Pembahasan Soal:

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Roboguru

Diketahui  dan . Nilai

Pembahasan Soal:

Dengan menerapkan sifat-sifat limit fungsi pada pengurangan dan pemangkatan fungsi, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 5 of space open parentheses f squared open parentheses x close parentheses minus g squared open parentheses x close parentheses close parentheses end cell equals cell limit as x rightwards arrow 5 of space f squared open parentheses x close parentheses minus limit as x rightwards arrow 5 of space g squared open parentheses x close parentheses end cell row blank equals cell open square brackets limit as x rightwards arrow 5 of space f open parentheses x close parentheses close square brackets squared minus open square brackets limit as x rightwards arrow 5 of space g open parentheses x close parentheses close square brackets squared end cell row blank equals cell open parentheses 2 close parentheses squared minus open parentheses negative 1 close parentheses squared end cell row blank equals cell 4 minus 1 end cell row blank equals 3 end table end style 

Jadi, nilai begin mathsize 14px style limit as x rightwards arrow 5 of space open parentheses f squared open parentheses x close parentheses minus g squared open parentheses x close parentheses close parentheses end style adalah 3.

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Roboguru

Jika nilai dari adalah ....

Pembahasan Soal:

Mencari f(x+h)

table attributes columnalign right center left columnspacing 0px end attributes row cell straight f open parentheses straight x plus straight h close parentheses end cell equals cell open parentheses straight x plus straight h close parentheses squared minus 2 open parentheses straight x plus straight h close parentheses minus 13 end cell row blank equals cell open parentheses straight x plus straight h close parentheses open parentheses straight x plus straight h close parentheses minus 2 straight x minus 2 straight h minus 13 end cell row blank equals cell straight x squared plus xh plus xh plus straight h squared minus 2 straight x minus 2 straight h minus 13 end cell row blank equals cell straight x squared plus 2 xh plus straight h squared minus 2 straight x minus 2 straight h minus 13 end cell end table

Subsitusi ke nilai lim

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight h rightwards arrow 0 of fraction numerator straight f open parentheses straight x plus straight h close parentheses minus straight f open parentheses straight x close parentheses over denominator straight h end fraction end cell equals cell limit as h rightwards arrow 0 of fraction numerator open parentheses x squared plus 2 x h plus h squared minus 2 x minus 2 h minus 13 close parentheses minus open parentheses x squared minus 2 x minus 13 close parentheses over denominator h end fraction end cell row blank equals cell limit as h rightwards arrow 0 of fraction numerator x squared plus 2 x h plus h squared minus 2 x minus 2 h minus 13 minus x squared plus 2 x plus 13 over denominator h end fraction end cell row blank equals cell limit as h rightwards arrow 0 of fraction numerator 2 x h plus h squared minus 2 h over denominator h end fraction end cell row blank equals cell limit as h rightwards arrow 0 of 2 x plus h minus 2 end cell row blank equals cell 2 x minus 2 end cell end table

Jadi hasil dari straight f left parenthesis straight x right parenthesis equals straight x squared minus 2 straight x minus 13 nilai dari limit as straight h rightwards arrow 0 of fraction numerator straight f open parentheses straight x plus straight h close parentheses minus straight f open parentheses straight x close parentheses over denominator straight h end fraction= 2x -2

 

 

 

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