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Nilai dari  adalah ....

Pertanyaan

Nilai dari begin mathsize 14px style limit as x rightwards arrow negative 3 of invisible function application open parentheses fraction numerator x squared minus 9 over denominator x plus 3 end fraction close parentheses end style adalah ....

  1. begin mathsize 14px style negative 6 end style

  2. begin mathsize 14px style negative 3 end style

  3. begin mathsize 14px style 3 end style

  4. begin mathsize 14px style 6 end style

  5. begin mathsize 14px style 9 end style

Pembahasan Soal:

Misal begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator x squared minus 9 over denominator x plus 3 end fraction end style . Perhatikan bahwa pada saat begin mathsize 14px style x equals negative 3 end style, didapat begin mathsize 14px style f open parentheses negative 3 close parentheses equals fraction numerator open parentheses negative 3 close parentheses squared minus 9 over denominator negative 3 plus 3 end fraction equals 0 over 0 end style 

Oleh karena itu, pada saat begin mathsize 14px style x equals negative 3 end style, nilai begin mathsize 14px style f left parenthesis x right parenthesis end style tak tentu.

Kemudian, perhatikan tabel berikut ini!

Dari arah kiri, jika nilai begin mathsize 14px style x end style makin mendekati undefined, dapat diperhatikan bahwa nilai begin mathsize 14px style f left parenthesis x right parenthesis end style makin mendekati undefined.

Oleh karena itu,limit as x rightwards arrow negative 3 to the power of minus of invisible function application open parentheses fraction numerator x squared minus 9 over denominator x plus 3 end fraction close parentheses equals negative 6.

Dari arah kanan, jika nilai begin mathsize 14px style x end style makin mendekati undefined, dapat diperhatikan bahwa nilai begin mathsize 14px style f left parenthesis x right parenthesis end style makin mendekati undefined.

Oleh karena itu, limit as x rightwards arrow negative 3 to the power of plus of invisible function application open parentheses fraction numerator x squared minus 9 over denominator x plus 3 end fraction close parentheses equals negative 6.

Karena begin mathsize 14px style limit as x rightwards arrow negative 3 to the power of minus of invisible function application open parentheses fraction numerator x squared minus 9 over denominator x plus 3 end fraction close parentheses equals limit as x rightwards arrow negative 3 to the power of plus of invisible function application open parentheses fraction numerator x squared minus 9 over denominator x plus 3 end fraction close parentheses equals negative 6 end style , maka size 14px lim with size 14px x size 14px rightwards arrow size 14px minus size 14px 3 below begin mathsize 14px style open parentheses fraction numerator x squared minus 9 over denominator x plus 3 end fraction close parentheses end style size 14px equals size 14px minus size 14px 6.
 

Jadi, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Indah

Mahasiswa/Alumni Universitas Lampung

Terakhir diupdate 09 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai dari  adalah ....

Pembahasan Soal:

Misal begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator x squared minus 4 over denominator x minus 2 end fraction end style. Perhatikan bahwa pada saat begin mathsize 14px style x equals 2 end style, didapat hasil sebagai berikut.

begin mathsize 14px style f open parentheses 2 close parentheses equals fraction numerator 2 squared minus 4 over denominator 2 minus 2 end fraction equals 0 over 0 end style 

Jadi, pada saat begin mathsize 14px style x equals 2 end style, nilai begin mathsize 14px style f left parenthesis x right parenthesis end style tak tentu.

Kemudian, perhatikan tabel berikut ini!

Dari arah kiri, jika nilai begin mathsize 14px style x end style makin mendekati undefined, dapat diperhatikan bahwa nilai begin mathsize 14px style f left parenthesis x right parenthesis end style makin mendekati undefined.

Dengan demikian, didapat perhitungan sebagai berikut.

begin mathsize 14px style limit as x rightwards arrow 2 to the power of minus of open parentheses fraction numerator x squared minus 4 over denominator x minus 2 end fraction close parentheses equals 4 end style 

Dari arah kanan,  jika nilai begin mathsize 14px style x end style makin mendekati undefined, dapat diperhatikan bahwa nilai begin mathsize 14px style f left parenthesis x right parenthesis end style juga makin mendekati undefined.

Dengan demikian, didapat perhitungan sebagai berikut.

begin mathsize 14px style limit as x rightwards arrow 2 to the power of plus of open parentheses fraction numerator x squared minus 4 over denominator x minus 2 end fraction close parentheses equals 4 end style

 

Karena begin mathsize 14px style limit as x rightwards arrow 2 to the power of minus of invisible function application open parentheses fraction numerator x squared minus 4 over denominator x minus 2 end fraction close parentheses equals limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses fraction numerator x squared minus 4 over denominator x minus 2 end fraction close parentheses equals 4 end style, maka begin mathsize 14px style limit as x rightwards arrow 2 of invisible function application open parentheses fraction numerator x squared minus 4 over denominator x minus 2 end fraction close parentheses equals 4 end style.

Hal ini juga dapat dilihat dari perhitungan berikut.

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 minus 2 end fraction close parentheses end cell equals cell limit as x rightwards arrow 2 of open parentheses fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis x plus 2 right parenthesis over denominator x minus 2 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 2 of open parentheses x plus 2 close parentheses end cell row blank equals cell 2 plus 2 end cell row blank equals 4 end table end style

Dengan demikian, didapat hasil perhitungannya sama dengan saat dilakukan pendekatan mendekati 2 dari kiri dan kanan.

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Didefinisikan fungsi  sebagai berikut. Jika ada, maka nilai dari −3⋅f(1) adalah ....

Pembahasan Soal:

Ingat bahwa supaya undefined ada, maka begin mathsize 14px style limit as x rightwards arrow 1 to the power of minus of invisible function application f open parentheses x close parentheses end style dan begin mathsize 14px style limit as x rightwards arrow 1 to the power of plus of invisible function application f open parentheses x close parentheses end style haruslah ada danbegin mathsize 14px style limit as x rightwards arrow 1 to the power of minus of invisible function application f open parentheses x close parentheses equals limit as x rightwards arrow 1 to the power of plus of invisible function application f open parentheses x close parentheses end style.

Dapat diperhatikan bahwa undefined berarti nilai limit dari f left parenthesis x right parenthesis ketika nilai x mendekati 1 dari arah kiri. Artinya, begin mathsize 14px style x less than 1 end style

Oleh karena itu, digunakan begin mathsize 14px style f open parentheses x close parentheses equals x squared end style. Akibatnya, didapat perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 to the power of minus of invisible function application f open parentheses x close parentheses end cell equals cell limit as x rightwards arrow 1 to the power of minus of invisible function application x squared end cell row blank equals cell 1 squared end cell row blank equals 1 end table end style

Kemudian, dapat diperhatikan bahwa undefined berarti nilai limit dari f left parenthesis x right parenthesis ketika nilai x mendekati 1 dari arah kanan. Artinya, begin mathsize 14px style x greater than 1 end style.

Oleh karena itu, digunakan begin mathsize 14px style f left parenthesis x right parenthesis equals m x plus 2 end style. Akibatnya, didapat perhitungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 to the power of plus of invisible function application f open parentheses x close parentheses end cell equals cell limit as x rightwards arrow 1 to the power of plus of invisible function application open parentheses m x plus 2 close parentheses end cell row blank equals cell open parentheses m times 1 plus 2 close parentheses end cell row blank equals cell m plus 2 end cell end table end style

Diketahui bahwa undefined ada, maka begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 1 to the power of minus of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank invisible function application end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank f end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell left parenthesis x right parenthesis 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 cell limit as x rightwards arrow 1 to the power of plus of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank invisible function application end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank f end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell left parenthesis x right parenthesis end cell end table end style.

Akibatnya, didapat perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 to the power of minus of invisible function application f open parentheses x close parentheses end cell equals cell limit as x rightwards arrow 1 to the power of plus of invisible function application f open parentheses x close parentheses end cell row 1 equals cell m plus 2 end cell row m equals cell negative 1 end cell end table end style

Oleh karena itu, nilai dari negative 3 times f open parentheses 1 close parentheses dapat dihitung sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 times f open parentheses 1 close parentheses end cell equals cell negative 3 open parentheses negative m close parentheses end cell row blank equals cell negative 3 open parentheses negative open parentheses negative 1 close parentheses close parentheses end cell row blank equals cell negative 3 end cell end table

Dengan demikian, nilai dari begin mathsize 14px style f open parentheses 1 close parentheses end style adalah negative 3.

Oleh karena itu, jawaban yang tepat adalah A.

1

Roboguru

Didefinisikan fungsi f(x) sebagai berikut.   Nilai dari  adalah ....

Pembahasan Soal:

Grafik fungsi fx dapat digambarkan sebagai berikut

Perhatikan nilai fungsi f(x) ketika nilai x mendekati 1.

  

Dari arah kiri, ketika nilai x mendekati 1, nilai f(x) mendekati -1. Jadi, didapat

begin mathsize 14px style limit as x rightwards arrow 1 to the power of minus of invisible function application f open parentheses x close parentheses equals negative 1 end style 

Kemudian, perhatikan dari arah kanan. Ketika nilai x mendekati 1, nilai f(x) mendekati 2. Jadi, didapat

begin mathsize 14px style limit as x rightwards arrow 1 to the power of plus of invisible function application f open parentheses x close parentheses equals 2 end style 

Karena begin mathsize 14px style limit as x rightwards arrow 1 to the power of minus of invisible function application f open parentheses x close parentheses not equal to limit as x rightwards arrow 1 to the power of plus of invisible function application f open parentheses x close parentheses end style, maka nilai dari undefined   tidak ada.

 

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Nilai dari  adalah ....

Pembahasan Soal:

Misal begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator 1 over denominator x minus 2 end fraction end style.

Pada saat begin mathsize 14px style x equals 2 end style, didapat perhitungan sebagai berikut.

begin mathsize 14px style f open parentheses 2 close parentheses equals fraction numerator 1 over denominator 2 minus 2 end fraction equals 1 over 0 end style 

Perhatikan bahwa pada saat begin mathsize 14px style x equals 2 end style, nilai begin mathsize 14px style f left parenthesis x right parenthesis end style menjadi tak terdefinisi.

Kemudian, perhatikan tabel berikut ini!

Perhatikan jika nilai begin mathsize 14px style x end style mendekati begin mathsize 14px style 2 end style dari arah kiri, nilai begin mathsize 14px style f left parenthesis x right parenthesis end style makin lama makin kecil menuju begin mathsize 14px style negative infinity end style. Jadi, didapat limit dari kiri sebagai berikut.

begin mathsize 14px style limit as x rightwards arrow 2 to the power of minus of invisible function application open parentheses fraction numerator 1 over denominator x minus 2 end fraction close parentheses equals negative infinity end style 

Perhatikan jika nilai begin mathsize 14px style x end style mendekati begin mathsize 14px style 2 end style dari arah kanan, nilai begin mathsize 14px style f left parenthesis x right parenthesis end style makin lama makin besar menuju undefined. Jadi, didapat limit dari kanan sebagai berikut.

begin mathsize 14px style limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses fraction numerator 1 over denominator x minus 2 end fraction close parentheses equals infinity end style 

Karena begin mathsize 14px style limit as x rightwards arrow 2 to the power of minus of invisible function application open parentheses fraction numerator 1 over denominator x minus 2 end fraction close parentheses not equal to limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses fraction numerator 1 over denominator x minus 2 end fraction close parentheses end style, maka undefined tidak ada.

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Nilai dari  adalah ....

Pembahasan Soal:

Perhatikan bahwa

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 invisible function application fraction numerator open parentheses 3 minus square root of 7 plus x end root close parentheses open parentheses square root of 6 minus x end root plus 2 close parentheses over denominator x minus 2 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of invisible function application open parentheses fraction numerator open parentheses 3 minus square root of 7 plus x end root close parentheses open parentheses square root of 6 minus x end root plus 2 close parentheses over denominator x minus 2 end fraction times fraction numerator 3 plus square root of 7 plus x end root over denominator 3 plus square root of 7 plus x end root end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 2 of invisible function application fraction numerator open parentheses 3 minus square root of 7 plus x end root close parentheses open parentheses 3 plus square root of 7 plus x end root close parentheses open parentheses square root of 6 minus x end root plus 2 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses 3 plus square root of 7 plus x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of invisible function application fraction numerator open parentheses 9 minus open parentheses 7 plus x close parentheses close parentheses open parentheses square root of 6 minus x end root plus 2 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses 3 plus square root of 7 plus x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of invisible function application fraction numerator open parentheses 2 minus x close parentheses open parentheses square root of 6 minus x end root plus 2 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses 3 plus square root of 7 plus x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of invisible function application fraction numerator negative open parentheses x minus 2 close parentheses open parentheses square root of 6 minus x end root plus 2 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses 3 plus square root of 7 plus x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of invisible function application fraction numerator negative open parentheses square root of 6 minus x end root plus 2 close parentheses over denominator 3 plus square root of 7 plus x end root end fraction end cell row blank equals cell fraction numerator negative open parentheses square root of 6 minus 2 end root plus 2 close parentheses over denominator 3 plus square root of 7 plus 2 end root end fraction end cell row blank equals cell fraction numerator negative open parentheses square root of 4 plus 2 close parentheses over denominator 3 plus square root of 9 end fraction end cell row blank equals cell fraction numerator negative open parentheses 2 plus 2 close parentheses over denominator 3 plus 3 end fraction end cell row blank equals cell negative 4 over 6 end cell row blank equals cell negative 2 over 3 end cell end table end style 

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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