Roboguru

Pertanyaan

limit as x rightwards arrow 0 of fraction numerator x squared minus 1 over denominator x squared minus x end fraction equals....   

Pembahasan Soal:

Dengan cara pemfaktoran.

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

Karena, hasil limit kiri dan kanan menedekati tak hingga, maka limit tersebut tidak ada.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Nur

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Terakhir diupdate 01 Mei 2021

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Pertanyaan yang serupa

Nilai  adalah ...

Pembahasan Soal:

Nilai limit fungsi tersebut adalah sebagai berikut.

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

Oleh karena itu, jawaban yang tepat adalah D.

0

Roboguru

Hitunglah nilai Limit berikut! a.

Pembahasan Soal:

Dengan menggunakan metode pemfaktoran, maka didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator x squared minus 3 x plus 2 over denominator x squared minus 4 end fraction end cell equals cell limit as x rightwards arrow 2 of fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 2 right parenthesis left parenthesis x plus 2 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator left parenthesis x minus 1 right parenthesis over denominator left parenthesis x plus 2 right parenthesis end fraction end cell row blank equals cell fraction numerator 2 minus 1 over denominator 2 plus 2 end fraction end cell row blank equals cell 1 fourth end cell end table

Jadi, nilai limit tersebut adalah 1 fourth.

0

Roboguru

Tentukan nilai dari

Pembahasan Soal:

Perhatikan perhitungan berikut.

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 fraction numerator 6 over denominator x squared minus x minus 2 end fraction minus fraction numerator 2 over denominator x minus 2 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator 6 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x plus 1 right parenthesis end fraction minus fraction numerator 2 over denominator left parenthesis x minus 2 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator 6 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x plus 1 right parenthesis end fraction minus fraction numerator 2 left parenthesis x plus 1 right parenthesis over denominator left parenthesis x minus 2 right parenthesis left parenthesis x plus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator 6 minus 2 open parentheses x plus 1 close parentheses over denominator left parenthesis x minus 2 right parenthesis left parenthesis x plus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator 6 minus 2 x minus 2 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x plus 1 right parenthesis end fraction end cell row blank blank cell stack equals lim with x rightwards arrow 2 below fraction numerator negative 2 x plus 4 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x plus 1 right parenthesis end fraction end cell row blank blank cell stack equals lim with x rightwards arrow 2 below fraction numerator negative 2 left parenthesis x minus 2 right parenthesis over denominator left parenthesis x minus 2 right parenthesis left parenthesis x plus 1 right parenthesis end fraction end cell row blank blank cell stack equals lim with x rightwards arrow 2 below fraction numerator negative 2 over denominator left parenthesis x plus 1 right parenthesis end fraction times fraction numerator left parenthesis x minus 2 right parenthesis over denominator left parenthesis x minus 2 right parenthesis end fraction end cell row blank blank cell stack equals lim with x rightwards arrow 2 below fraction numerator negative 2 over denominator left parenthesis x plus 1 right parenthesis end fraction times 1 end cell row blank equals cell fraction numerator negative 2 over denominator 2 plus 1 end fraction times 1 end cell row blank equals cell negative 2 over 3 end cell end table end style

Jadi, diperoleh begin mathsize 14px style limit as x rightwards arrow 2 of fraction numerator 6 over denominator x squared minus x minus 2 end fraction minus fraction numerator 2 over denominator x minus 2 end fraction equals negative 2 over 3 end style.

0

Roboguru

Hitunglah nilai limit fungsi berikut!

Pembahasan Soal:

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

undefined 

Jadi, nilai dari begin mathsize 14px style limit as x rightwards arrow 3 of space fraction numerator 3 x squared minus 4 x minus 15 over denominator x squared plus x minus 12 end fraction end style adalah begin mathsize 14px style 2 end style.

0

Roboguru

Pembahasan Soal:

Diketahui begin mathsize 14px style limit as x rightwards arrow 4 of space fraction numerator x squared minus 9 x plus 20 over denominator x minus 4 end fraction end style maka untuk menentukan nilainya substitusi x=4 ke dalam x4x29x+20 sehingga

limx4x4x29x+20===44(4)29(4)+2001636+2000

Karena hasilnya 00 maka dapat dilakukan cara pemfaktoran untuk menentukan nilai limitnya sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space fraction numerator x squared minus 9 x plus 20 over denominator x minus 4 end fraction space end cell equals cell space limit as x rightwards arrow 4 of space fraction numerator open parentheses x minus 5 close parentheses open parentheses x minus 4 close parentheses over denominator x minus 4 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of open parentheses x minus 5 close parentheses end cell row blank equals cell 4 minus 5 end cell row blank equals cell negative 1 end cell end table end style   

Dengan demikian, begin mathsize 14px style limit as x rightwards arrow 4 of space table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x squared minus 9 x plus 20 over denominator x minus 4 end fraction 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 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 1 end table end style.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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