Roboguru

Nilai

Pertanyaan

Nilai limit as x rightwards arrow negative 1 of invisible function application fraction numerator x squared minus 1 over denominator x minus 1 end fraction equals horizontal ellipsis

  1. negative 8

  2. negative 6

  3. negative 4

  4. 0

  5. 2

Pembahasan Soal:

Ingat cara menentukan nilai sebuah limit dengan cara pemfaktoran dari persamaan yang ada.

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

Jadi, dapat disimpulkan bahwa nilai dari limit as x rightwards arrow negative 1 of invisible function application fraction numerator x squared minus 1 over denominator x minus 1 end fraction adalah 0.

Oleh karena itu, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Iqbal

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 05 Juni 2021

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Pertanyaan yang serupa

Nilai dari .

Pembahasan Soal:

0

Roboguru

Nilai dari ...

Pembahasan Soal:

Untuk menentukan nilai limit fungsi tersebut salah satunya dapat menggunakan metode pemfaktoran sehingga:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 1 of fraction numerator 3 x squared minus x minus 4 over denominator x squared minus 1 end fraction end cell equals cell limit as x rightwards arrow negative 1 of fraction numerator open parentheses 3 x minus 4 close parentheses open parentheses x plus 1 close parentheses over denominator open parentheses x minus 1 close parentheses open parentheses x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 1 of fraction numerator 3 x minus 4 over denominator x minus 1 end fraction end cell row blank equals cell fraction numerator 3 open parentheses negative 1 close parentheses minus 4 over denominator negative 1 minus 1 end fraction end cell row blank equals cell fraction numerator negative 7 over denominator negative 2 end fraction end cell row blank equals cell 7 over 2 end cell end table end style 

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Nilai

Pembahasan Soal:

Limit dapat diartikan sebagai menuju suatu batas, sesuatu yang dekat namun tidak dapat dicapai. Cara penyelesaian nilai undefined mendekati berhingga adalah dengan substitusi, pemfaktoran, dan dikalikan dengan sekawannya.

Persoalan di atas dapat digunakan metode substitusi.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator 3 x plus 4 x to the power of negative 1 end exponent over denominator 4 x minus x to the power of negative 1 end exponent end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator 3 x squared plus 4 over denominator 4 x squared minus 1 end fraction end cell row blank equals cell fraction numerator 3 open parentheses 0 close parentheses squared plus 4 over denominator 4 open parentheses 0 close parentheses squared minus 1 end fraction end cell row blank equals cell fraction numerator 4 over denominator negative 1 end fraction end cell row blank equals cell negative 4 end cell end table end style 

Jadi, diperoleh begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator 3 x plus 4 x to the power of negative 1 end exponent over denominator 4 x minus x to the power of negative 1 end exponent end fraction equals negative 4 end style.

0

Roboguru

Hitunglah nilai limit berikut ini.

Pembahasan Soal:

Nilai dari begin mathsize 14px style limit as x rightwards arrow 2 of fraction numerator 2 x squared minus 5 x plus 2 over denominator x minus 2 end fraction end style dapat ditentukan dengan menggunakan metode pemfaktoran.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator 2 x squared minus 5 x plus 2 over denominator x minus 2 end fraction end cell equals cell limit as x rightwards arrow 2 of fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses x minus 2 close parentheses over denominator x minus 2 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of open parentheses 2 x minus 1 close parentheses end cell row blank equals cell 2 open parentheses 2 close parentheses minus 1 end cell row blank equals 3 end table end style

Dengan demikian, nilai begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 2 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 x squared minus 5 x plus 2 over denominator x minus 2 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 3 end table end style

0

Roboguru

Pembahasan Soal:

Perhatikan perhitungan berikut!

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator x to the power of 7 minus 2 x to the power of 5 plus 1 over denominator x cubed minus 3 x squared plus 2 end fraction end cell equals cell fraction numerator 1 to the power of 7 minus 2 times 1 to the power of 5 plus 1 over denominator 1 cubed minus 3 times 1 squared plus 2 end fraction end cell row blank equals cell fraction numerator 1 minus 2 times 1 plus 1 over denominator 1 minus 3 times 1 plus 2 end fraction end cell row blank equals cell fraction numerator 1 minus 2 plus 1 over denominator 1 minus 3 plus 2 end fraction end cell row blank equals cell fraction numerator negative 1 plus 1 over denominator negative 2 plus 2 end fraction end cell row blank equals cell 0 over 0 end cell end table 

Karena hasil dari limit tersebut ketika disubstitusikan x equals 1 merupakan bilangan tak tentu 0 over 0, maka kita lakukan pemfaktoran pada bentuk limit tersebut sebagai berikut:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 1 of space fraction numerator straight x to the power of 7 minus 2 straight x to the power of 5 plus 1 over denominator straight x cubed minus 3 straight x squared plus 2 end fraction end cell equals cell limit as straight x rightwards arrow 1 of space fraction numerator straight x to the power of 7 minus 2 straight x to the power of 5 plus 1 over denominator straight x cubed minus 3 straight x squared plus 2 end fraction space colon space fraction numerator straight x to the power of 6 plus straight x to the power of 5 minus straight x to the power of 4 minus straight x cubed minus straight x squared minus straight x minus 1 over denominator straight x squared minus 2 straight x minus 2 end fraction end cell row cell 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 end cell equals cell limit as straight x rightwards arrow 1 of space fraction numerator straight x to the power of 6 plus straight x to the power of 5 minus straight x to the power of 4 minus straight x cubed minus straight x squared minus straight x minus 1 over denominator straight x squared minus 2 straight x minus 2 end fraction end cell row cell 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 end cell equals cell fraction numerator 1 to the power of 6 plus 1 to the power of 5 minus 1 to the power of 4 minus 1 cubed minus 1 squared minus 1 minus 1 over denominator 1 squared minus 2 open parentheses 1 close parentheses minus 2 end fraction end cell row cell 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 end cell equals cell fraction numerator 1 plus 1 minus 1 minus 1 minus 1 minus 1 minus 1 over denominator 1 minus 2 minus 2 end fraction end cell row cell 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 end cell equals cell fraction numerator negative 3 over denominator negative 3 end fraction end cell row cell 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 end cell equals 1 end table end style 

Jadi, limit as x rightwards arrow 1 of fraction numerator x to the power of 7 minus 2 x to the power of 5 plus 1 over denominator x cubed minus 3 x squared plus 2 end fraction equals 1.

0

Roboguru

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