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Pertanyaan

limit as x rightwards arrow 2 of fraction numerator x squared minus 2 x over denominator x minus 2 end fraction equals space...

Pembahasan Soal:

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

Dengan demikian, hasil dari limit as x rightwards arrow 2 of fraction numerator x squared minus 2 x over denominator x minus 2 end fraction adalah 2. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Indah

Mahasiswa/Alumni Universitas Diponegoro

Terakhir diupdate 04 Juni 2021

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Pertanyaan yang serupa

Tentukan nilai dari soal di bawah ini 3.

Pembahasan Soal:

substitusikan nilai undefined ke fungsi sehingga diperoleh
begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator 15 x to the power of 4 plus 6 x squared plus 10 over denominator 5 minus 7 x minus 10 x squared end fraction equals 10 over 5 equals 2 end style 

0

Roboguru

Hasil dari  adalah ...

Pembahasan Soal:

Bila kita gunakan substitusi maka kita akan mendapatkan bentuk tak tentu. Oleh karena itu, kita lakukan dengan cara pemfaktoran.

limit as x rightwards arrow 2 of fraction numerator 3 x squared minus 4 x minus 4 over denominator x squared minus 4 end fraction equals limit as x rightwards arrow 2 of fraction numerator open parentheses 3 x plus 2 close parentheses open parentheses x minus 2 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses x plus 2 close parentheses 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 equals limit as x rightwards arrow 2 of fraction numerator open parentheses 3 x plus 2 close parentheses over denominator open parentheses x plus 2 close parentheses 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 equals fraction numerator 3.2 plus 2 over denominator 2 plus 2 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 equals 8 over 4 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 2 

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Tentukan nilai dari !

Pembahasan Soal:

Diketahui begin mathsize 14px style limit as x rightwards arrow 1 of fraction numerator 3 x minus 2 over denominator x squared minus 1 end fraction end style. Maka:

begin mathsize 14px style limit as x rightwards arrow 1 of equals fraction numerator 3 x minus 2 over denominator x squared minus 1 end fraction equals fraction numerator 3 open parentheses 1 close parentheses minus 2 over denominator 1 squared minus 1 end fraction equals fraction numerator 3 minus 2 over denominator 1 minus 1 end fraction equals 1 over 0 end style 

Dengan bentuk yang demikian, limit tersebut tidak dapat diselesaikan. Sehingga harus ada perubahan dari soal, yaitu:

begin mathsize 14px style limit as x rightwards arrow 1 of fraction numerator 3 x minus 3 over denominator x squared minus 1 end fraction equals limit as x rightwards arrow 1 of fraction numerator 3 open parentheses x minus 1 close parentheses over denominator open parentheses x minus 1 close parentheses open parentheses x plus 1 close parentheses end fraction limit as x rightwards arrow 1 of fraction numerator 3 x minus 3 over denominator x squared minus 1 end fraction equals limit as x rightwards arrow 1 of fraction numerator 3 over denominator open parentheses x plus 1 close parentheses end fraction limit as x rightwards arrow 1 of fraction numerator 3 x minus 3 over denominator x squared minus 1 end fraction equals fraction numerator 3 over denominator open parentheses 1 plus 1 close parentheses end fraction limit as x rightwards arrow 1 of fraction numerator 3 x minus 3 over denominator x squared minus 1 end fraction equals 3 over 2 end style 

atau

 begin mathsize 14px style limit as x rightwards arrow 1 of fraction numerator 2 x minus 2 over denominator x squared minus 1 end fraction equals limit as x rightwards arrow 1 of fraction numerator 2 open parentheses x minus 1 close parentheses over denominator open parentheses x minus 1 close parentheses open parentheses x plus 1 close parentheses end fraction limit as x rightwards arrow 1 of fraction numerator 3 x minus 3 over denominator x squared minus 1 end fraction equals limit as x rightwards arrow 1 of fraction numerator 2 over denominator open parentheses x plus 1 close parentheses end fraction limit as x rightwards arrow 1 of fraction numerator 3 x minus 3 over denominator x squared minus 1 end fraction equals fraction numerator 2 over denominator 1 plus 1 end fraction limit as x rightwards arrow 1 of fraction numerator 3 x minus 3 over denominator x squared minus 1 end fraction equals 2 over 2 limit as x rightwards arrow 1 of fraction numerator 3 x minus 3 over denominator x squared minus 1 end fraction equals 1 end style 

Dengan demikian, nilai dari limit tersebut adalah begin mathsize 14px style 3 over 2 space atau space 1 end style.

0

Roboguru

Pembahasan Soal:

Untuk menyelesaikan bentuk limit diatas kita dapat menggunkan 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 open parentheses fraction numerator x squared minus 4 over denominator x squared minus 3 x plus 2 end fraction close parentheses end cell equals cell limit as x rightwards arrow 2 of fraction numerator open parentheses x minus 2 close parentheses open parentheses x plus 2 close parentheses over denominator open parentheses x minus 1 close parentheses open parentheses x minus 2 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator open parentheses x plus 2 close parentheses over denominator open parentheses x minus 1 close parentheses end fraction end cell row blank equals cell fraction numerator 2 plus 2 over denominator 2 minus 1 end fraction end cell row blank equals 4 end table end style

Jadi, jawaban yang tepat adalah E

0

Roboguru

Tentukan nilai limit berikut. b.

Pembahasan Soal:

Hasil substitusi nilai begin mathsize 14px style x end style ke fungsi begin mathsize 14px style f open parentheses x close parentheses end style adalah begin mathsize 14px style 0 over 0 end style sehingga penyelesaian menggunakan metode pemfaktoran sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of space fraction numerator 2 x squared minus 5 x plus 2 over denominator x squared minus 4 end fraction end cell equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses x minus 2 close parentheses over denominator open parentheses x plus 2 close parentheses open parentheses x minus 2 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator 2 x minus 1 over denominator x plus 2 end fraction end cell row blank equals cell fraction numerator 2 times 2 minus 1 over denominator 2 plus 2 end fraction end cell row blank equals cell 3 over 4 end cell end table end style

Dengan demikian, nilai begin mathsize 14px style limit as x rightwards arrow 2 of space fraction numerator 2 x squared minus 5 x plus 2 over denominator x squared minus 4 end fraction equals 3 over 4 end style

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