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Nilai dari  adalah ...

Pertanyaan

Nilai dari begin mathsize 14px style limit as straight x rightwards arrow 0 of fraction numerator left parenthesis straight x squared minus 1 right parenthesis left parenthesis straight x squared minus 4 straight x right parenthesis over denominator straight x squared minus straight x end fraction end style adalah ...

  1. begin mathsize 14px style negative 4 end style 

  2. size 14px minus size 14px 2 

  3. size 14px 0 

  4. size 14px 2 

  5. size 14px 4 

Pembahasan Soal:

Dengan menerapkan metode pemfaktoran, maka diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 0 of fraction numerator left parenthesis straight x squared minus 1 right parenthesis left parenthesis straight x squared minus 4 straight x right parenthesis over denominator straight x squared minus straight x end fraction end cell equals cell limit as straight x rightwards arrow 0 of fraction numerator left parenthesis straight x plus 1 right parenthesis left parenthesis straight x minus 1 right parenthesis times straight x left parenthesis straight x minus 4 right parenthesis over denominator straight x left parenthesis straight x minus 1 right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 0 of fraction numerator straight x times left parenthesis straight x plus 1 right parenthesis left parenthesis straight x minus 1 right parenthesis left parenthesis straight x minus 4 right parenthesis over denominator straight x left parenthesis straight x minus 1 right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 0 of left parenthesis straight x plus 1 right parenthesis left parenthesis straight x minus 4 right parenthesis end cell row blank equals cell left parenthesis 0 plus 1 right parenthesis left parenthesis 0 minus 4 right parenthesis end cell row blank equals cell 1 times left parenthesis negative 4 right parenthesis end cell row blank equals cell negative 4 end cell end table end style  

Jadi, Nilai begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow 0 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator left parenthesis straight x squared minus 1 right parenthesis left parenthesis straight x squared minus 4 straight x right parenthesis over denominator straight x squared minus straight x end fraction 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 4 end table end style 

Oleh karena itu, Jawaban yang benar adalah A

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. RGFLLIMA

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

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 

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 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.

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.

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 

Roboguru

Hitunglah nilai limit fungsi berikut!

Pembahasan Soal:

Dengan menerapkan metode pemfaktoran, limit fungsi di atas dapat diselesaikan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 7 of space fraction numerator x squared minus 49 over denominator x plus 7 end fraction end cell equals cell limit as x rightwards arrow negative 7 of space fraction numerator up diagonal strike open parentheses x plus 7 close parentheses end strike open parentheses x minus 7 close parentheses over denominator up diagonal strike x plus 7 end strike end fraction end cell row blank equals cell limit as x rightwards arrow negative 7 of space open parentheses x minus 7 close parentheses end cell row blank equals cell open parentheses negative 7 close parentheses minus 7 end cell row blank equals cell negative 14 end cell end table end style 

Jadi, nilai dari begin mathsize 14px style limit as x rightwards arrow negative 7 of space fraction numerator x squared minus 49 over denominator x plus 7 end fraction end style adalah begin mathsize 14px style negative 14 end style.

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