Roboguru

Nilai dari  adalah ....

Pertanyaan

Nilai dari begin mathsize 14px style limit as straight x rightwards arrow 7 of fraction numerator straight x minus 7 over denominator square root of straight x minus square root of 7 end fraction end style adalah ....

  1. begin mathsize 14px style 7 square root of 7 end style 

  2. 3 square root of size 14px 7 

  3. begin mathsize 14px style 2 square root of 7 end style  

  4. size 14px 1 over size 14px 7 square root of size 14px 7 

  5. size 14px 1 over size 14px 14 square root of size 14px 7 

Pembahasan Soal:

Dengan menerapkan konsep perkalian sekawan;

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 7 of fraction numerator straight x minus 7 over denominator square root of straight x minus square root of 7 end fraction end cell equals cell limit as straight x rightwards arrow 7 of fraction numerator left parenthesis straight x minus 7 right parenthesis times left parenthesis square root of straight x plus square root of 7 right parenthesis over denominator left parenthesis square root of straight x minus square root of 7 right parenthesis times left parenthesis square root of straight x plus square root of 7 right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 7 of fraction numerator left parenthesis straight x minus 7 right parenthesis times left parenthesis square root of straight x plus square root of 7 right parenthesis over denominator left parenthesis straight x minus 7 right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 7 of left parenthesis square root of straight x plus square root of 7 right parenthesis end cell row blank equals cell left parenthesis square root of 7 plus square root of 7 right parenthesis end cell row blank equals cell 2 square root of 7 end cell end table end style 

Jadi, Nilai Error converting from MathML to accessible text..

Oleh karena itu, Jawaban yang benar adalah C

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. RGFLLIMA

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

Nilai  adalah ....

Pembahasan Soal:

Dengan mengalikan akar sekawan.

begin mathsize 14px style limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator square root of 12 minus square root of x plus 8 end root end fraction equals limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator square root of 12 minus square root of x plus 8 end root end fraction times fraction numerator square root of 12 plus square root of x plus 8 end root over denominator square root of 12 plus square root of x plus 8 end root end fraction equals limit as x rightwards arrow 4 of fraction numerator open parentheses 4 minus x close parentheses open parentheses square root of 12 plus square root of x plus 8 end root close parentheses over denominator 12 minus open parentheses x plus 8 close parentheses end fraction equals limit as x rightwards arrow 4 of fraction numerator up diagonal strike open parentheses 4 minus x close parentheses end strike open parentheses square root of 12 plus square root of x plus 8 end root close parentheses over denominator up diagonal strike open parentheses 4 minus x close parentheses end strike end fraction equals limit as x rightwards arrow 4 of square root of 12 plus square root of x plus 8 end root equals square root of 12 plus square root of 4 plus 8 end root equals square root of 12 plus square root of 12 equals 2 square root of 12 equals 2 square root of 4 times 3 end root equals 2 times 2 square root of 3 equals 4 square root of 3 end style 

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Pembahasan Soal:

Limit begin mathsize 14px style f open parentheses x close parentheses end style mendekati undefined sama dengan undefined, ditulis:

begin mathsize 14px style limit as x rightwards arrow c of space f open parentheses x close parentheses equals L end style  

jika untuk setiap undefined yang cukup dekat dengan undefined tetapi begin mathsize 14px style x not equal to c end styleundefined mendekati undefined.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as h rightwards arrow 0 of space fraction numerator square root of 4 x plus 4 h end root minus square root of 4 x end root over denominator h end fraction end cell equals cell limit as h rightwards arrow 0 of space fraction numerator square root of 4 x plus 4 h end root minus square root of 4 x end root over denominator h end fraction cross times fraction numerator square root of 4 x plus 4 h end root plus square root of 4 x end root over denominator square root of 4 x plus 4 h end root plus square root of 4 x end root end fraction end cell row blank equals cell limit as h rightwards arrow 0 of space fraction numerator open parentheses 4 x plus 4 h close parentheses minus 4 x over denominator h open parentheses square root of 4 x plus 4 h end root plus square root of 4 x end root close parentheses end fraction end cell row blank equals cell limit as h rightwards arrow 0 of space fraction numerator 4 x plus 4 h minus 4 x over denominator h open parentheses square root of 4 x plus 4 h end root plus square root of 4 x end root close parentheses end fraction end cell row blank equals cell limit as h rightwards arrow 0 of space fraction numerator 4 h over denominator h open parentheses square root of 4 x plus 4 h end root plus square root of 4 x end root close parentheses end fraction end cell row blank equals cell limit as h rightwards arrow 0 of space fraction numerator 4 over denominator open parentheses square root of 4 x plus 4 h end root plus square root of 4 x end root close parentheses end fraction end cell row blank equals cell fraction numerator 4 over denominator open parentheses square root of 4 x plus 4 left parenthesis 0 right parenthesis end root plus square root of 4 x end root close parentheses end fraction end cell row blank equals cell fraction numerator 4 over denominator square root of 4 x end root plus square root of 4 x end root end fraction end cell row blank equals cell fraction numerator 4 over denominator 2 square root of 4 x end root end fraction end cell row blank equals cell fraction numerator 2 over denominator square root of 4 x end root end fraction end cell row blank equals cell fraction numerator 2 over denominator square root of 4 x end root end fraction cross times fraction numerator square root of 4 x end root over denominator square root of 4 x end root end fraction end cell row blank equals cell fraction numerator 2 square root of 4 x end root over denominator 4 x end fraction end cell row blank equals cell fraction numerator 2 cross times 2 square root of x over denominator 4 x end fraction end cell row blank equals cell fraction numerator 4 square root of x over denominator 4 x end fraction end cell row blank equals cell fraction numerator square root of x over denominator x end fraction end cell row blank blank blank end table end style 

0

Roboguru

Nilai dari  adalah...

Pembahasan Soal:

Dengan mengalikan fungsi dengan akar sekawan dari penyebut, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of fraction numerator x minus 4 over denominator square root of x minus 2 end fraction end cell equals cell limit as x rightwards arrow 4 of fraction numerator x minus 4 over denominator square root of x minus 2 end fraction cross times fraction numerator square root of x plus 2 over denominator square root of x plus 2 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator x minus 4 open parentheses square root of x plus 2 close parentheses over denominator x minus 4 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of square root of x plus 2 end cell row blank equals cell square root of 4 plus 2 end cell row blank equals cell 2 plus 2 end cell row blank equals 4 end table

Jadi, 

limit as x rightwards arrow 4 of fraction numerator x minus 4 over denominator square root of x minus 2 end fraction equals 4

0

Roboguru

Tentukan nilai

Pembahasan Soal:

begin mathsize 14px style limit as x rightwards arrow 3 of space fraction numerator x squared minus 9 over denominator square root of x squared plus 16 end root minus 5 end fraction equals limit as x rightwards arrow 3 of space fraction numerator x squared minus 9 over denominator square root of x squared plus 16 end root minus 5 end fraction times fraction numerator square root of x squared plus 16 end root plus 5 over denominator square root of x squared plus 16 end root plus 5 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 space equals limit as x rightwards arrow 3 of space fraction numerator open parentheses x squared minus 9 close parentheses open parentheses square root of x squared plus 16 end root plus 5 close parentheses over denominator x squared plus 16 minus 25 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 space equals limit as x rightwards arrow 3 of space fraction numerator left parenthesis x squared minus 9 right parenthesis open parentheses square root of x squared plus 16 end root plus 5 close parentheses over denominator x squared minus 9 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 space equals limit as x rightwards arrow 3 of space open parentheses square root of x squared plus 16 end root plus 5 close parentheses 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 space equals square root of left parenthesis 3 right parenthesis squared plus 16 end root plus 5 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 space equals square root of 9 plus 16 end root plus 5 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 space equals square root of 25 plus 5 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 space equals 5 plus 5 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 space equals 10 end style 

Jadi, begin mathsize 14px style limit as x rightwards arrow 3 of space fraction numerator x squared minus 9 over denominator square root of x squared plus 16 end root minus 5 end fraction equals 10 end style.

0

Roboguru

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row straight L equals cell fraction numerator straight b minus straight q over denominator 2 square root of straight a end fraction end cell row blank equals cell fraction numerator negative 2 minus left parenthesis negative 2 right parenthesis over denominator 2 square root of 1 end fraction end cell row blank equals cell 0 over 2 end cell row blank equals 0 end table

Jadi hasil dari limit as x rightwards arrow infinity of open parentheses square root of x squared minus 2 x minus 1 end root minus square root of x squared minus 2 x plus 1 end root close parentheses equals 0

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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