Roboguru

Nilai dari x→−3lim​x2+7x+12x+12​−3​=…

Pertanyaan

Nilai dari limit as x rightwards arrow negative 3 of space fraction numerator square root of x plus 12 end root minus 3 over denominator x squared plus 7 x plus 12 end fraction equals horizontal ellipsis 

  1. 0 

  2. 1 over 12 

  3. 1 over 8 

  4. 1 over 6 

  5. 1 fourth 

Pembahasan Soal:

Nilainya didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 3 of space fraction numerator square root of x plus 12 end root minus 3 over denominator x squared plus 7 x plus 12 end fraction end cell equals cell limit as x rightwards arrow negative 3 of space fraction numerator square root of x plus 12 end root minus 3 over denominator x squared plus 7 x plus 12 end fraction times fraction numerator square root of x plus 12 end root plus 3 over denominator square root of x plus 12 end root plus 3 end fraction end cell row blank equals cell limit as x rightwards arrow negative 3 of space fraction numerator x plus 12 minus 9 over denominator open parentheses x squared plus 7 x plus 12 close parentheses open parentheses square root of x plus 12 end root plus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 3 of space fraction numerator up diagonal strike x plus 3 end strike over denominator up diagonal strike open parentheses x plus 3 close parentheses end strike open parentheses x plus 4 close parentheses open parentheses square root of x plus 12 end root plus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 3 of space fraction numerator 1 over denominator open parentheses x plus 4 close parentheses open parentheses square root of x plus 12 end root plus 3 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator open parentheses open parentheses negative 3 close parentheses plus 4 close parentheses open parentheses square root of open parentheses negative 3 close parentheses plus 12 end root plus 3 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator open parentheses 1 close parentheses open parentheses 3 plus 3 close parentheses end fraction end cell row blank equals cell 1 over 6 end cell end table  

Dengan demikian, nilai dari limit as x rightwards arrow negative 3 of space fraction numerator square root of x plus 12 end root minus 3 over denominator x squared plus 7 x plus 12 end fraction equals 1 over 6.

Jadi, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Endah

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Jika U(x)=2x2+5x−7​ danV(x)=3x2+8x−25​, maka nilai dari x→3lim​x−3U(x)−V(x)​ adalah...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator U open parentheses x close parentheses minus V open parentheses x close parentheses over denominator x minus 3 end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator square root of 2 x squared plus 5 x minus 7 end root minus square root of 3 x squared plus 8 x minus 25 end root over denominator x minus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator square root of 2 x squared plus 5 x minus 7 end root minus square root of 3 x squared plus 8 x minus 25 end root over denominator x minus 3 end fraction cross times fraction numerator square root of 2 x squared plus 5 x minus 7 end root plus square root of 3 x squared plus 8 x minus 25 end root over denominator square root of 2 x squared plus 5 x minus 7 end root plus square root of 3 x squared plus 8 x minus 25 end root end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 2 x squared plus 5 x minus 7 close parentheses minus open parentheses 3 x squared plus 8 x minus 25 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses square root of 2 x squared plus 5 x minus 7 end root plus square root of 3 x squared plus 8 x minus 25 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator negative x squared minus 3 x plus 18 over denominator open parentheses x minus 3 close parentheses open parentheses square root of 2 x squared plus 5 x minus 7 end root plus square root of 3 x squared plus 8 x minus 25 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator negative open parentheses x squared plus 3 x minus 18 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses square root of 2 x squared plus 5 x minus 7 end root plus square root of 3 x squared plus 8 x minus 25 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator negative open parentheses x minus 3 close parentheses open parentheses x plus 6 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses square root of 2 x squared plus 5 x minus 7 end root plus square root of 3 x squared plus 8 x minus 25 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator negative open parentheses x plus 6 close parentheses over denominator open parentheses square root of 2 x squared plus 5 x minus 7 end root plus square root of 3 x squared plus 8 x minus 25 end root close parentheses end fraction end cell row blank equals cell fraction numerator negative open parentheses 3 plus 6 close parentheses over denominator open parentheses square root of 2 times 3 squared plus 5 times 3 minus 7 end root plus square root of 3 times 3 squared plus 8 times 3 minus 25 end root close parentheses end fraction end cell row blank equals cell negative fraction numerator 9 over denominator square root of 26 plus square root of 26 end fraction end cell row blank equals cell negative fraction numerator 9 over denominator 2 square root of 26 end fraction end cell row blank equals cell negative 9 over 52 square root of 26 end cell end table end styleJadi, jawaban yang tepat adalah C.

0

Roboguru

Nilai x→1lim​2+2x​−6−2x​x3−x2​ adalah ...

Pembahasan Soal:

Substitusi x = 1:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator x cubed minus x squared over denominator square root of 2 plus 2 x end root minus square root of 6 minus 2 x end root end fraction end cell equals cell fraction numerator 1 cubed minus 1 squared over denominator square root of 2 plus 2 times 1 end root minus square root of 6 minus 2 times 1 end root end fraction end cell row blank equals cell fraction numerator 1 minus 1 over denominator square root of 4 minus square root of 4 end fraction end cell row blank equals cell 0 over 0 space open parentheses tak space tentu close parentheses end cell end table

Karena 0 over 0 merupakan bentuk tak tentu, maka tidak boleh. Jadi harus diselesaiakan dengan metode mengalikan akar sekawan: (ingat: open parentheses a plus b close parentheses open parentheses a minus b close parentheses equals a squared minus b squared)

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 1 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x cubed minus x squared over denominator square root of 2 plus 2 x end root minus square root of 6 minus 2 x end root end fraction end cell end table equals limit as x rightwards arrow 1 of fraction numerator x cubed minus x squared over denominator square root of 2 plus 2 x end root minus square root of 6 minus 2 x end root end fraction cross times fraction numerator square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root over denominator square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root end fraction equals limit as x rightwards arrow 1 of fraction numerator open parentheses x cubed minus x squared close parentheses open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator open parentheses square root of 2 plus 2 x end root close parentheses squared minus open parentheses square root of 6 minus 2 x end root close parentheses squared end fraction equals limit as x rightwards arrow 1 of fraction numerator open parentheses x cubed minus x squared close parentheses open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator left parenthesis 2 plus 2 x right parenthesis minus left parenthesis 6 minus 2 x right parenthesis end fraction equals limit as x rightwards arrow 1 of fraction numerator open parentheses x cubed minus x squared close parentheses open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator 2 plus 2 x minus 6 plus 2 x end fraction equals limit as x rightwards arrow 1 of fraction numerator open parentheses x cubed minus x squared close parentheses open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator 4 x minus 4 end fraction equals limit as x rightwards arrow 1 of fraction numerator x squared up diagonal strike open parentheses x minus 1 close parentheses end strike open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator 4 up diagonal strike left parenthesis x minus 1 right parenthesis end strike end fraction equals limit as x rightwards arrow 1 of fraction numerator x squared open parentheses square root of 2 plus 2 x end root plus square root of 6 minus 2 x end root close parentheses over denominator 4 end fraction equals fraction numerator 1 squared open parentheses square root of 2 plus 2 left parenthesis 1 right parenthesis end root plus square root of 6 minus 2 left parenthesis 1 right parenthesis end root close parentheses over denominator 4 end fraction equals fraction numerator square root of 4 plus square root of 4 over denominator 4 end fraction equals fraction numerator 2 plus 2 over denominator 4 end fraction equals 4 over 4 equals 1 

Jadi, jawaban yang tepat adalah D

0

Roboguru

Jika diketahui x→3lim​x2+16​−5ax2−9a​​=10 tentukan nilai a ?

Pembahasan Soal:

Soal di atas adalah bentuk limit tak tentu dan dapat diselesaikan dengan dalil l'hospital dengan rumus :

begin mathsize 14px style limit as straight x rightwards arrow straight a of fraction numerator straight f open parentheses straight x close parentheses over denominator straight g open parentheses straight x close parentheses end fraction equals limit as straight x rightwards arrow straight a of fraction numerator straight f apostrophe open parentheses straight x close parentheses over denominator straight g apostrophe open parentheses straight x close parentheses end fraction end style 

Maka kita dapat tentukan nilai a :

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell limit as straight x rightwards arrow 3 of fraction numerator ax squared minus 9 straight a over denominator square root of straight x squared plus 16 end root minus 5 end fraction end cell equals 10 row cell limit as straight x rightwards arrow 3 of fraction numerator ax squared minus 9 straight a over denominator open parentheses straight x squared plus 16 close parentheses to the power of begin display style 1 half end style end exponent minus 5 end fraction end cell equals 10 row cell limit as straight x rightwards arrow 3 of fraction numerator 2 ax over denominator begin display style 1 half end style open parentheses straight x squared plus 16 close parentheses to the power of begin display style negative 1 half end style end exponent times 2 straight x end fraction end cell equals 10 row cell fraction numerator 2 straight a open parentheses 3 close parentheses over denominator begin display style 1 half open parentheses open parentheses 3 close parentheses squared plus 16 close parentheses to the power of negative 1 half end exponent times 2 open parentheses 3 close parentheses end style end fraction end cell equals 10 row cell fraction numerator 6 straight a over denominator begin display style 1 half end style open parentheses 25 close parentheses to the power of negative begin display style 1 half end style end exponent times 6 end fraction end cell equals 10 row cell fraction numerator 6 straight a over denominator 3 times fraction numerator 1 over denominator square root of 25 end fraction end fraction end cell equals 10 row cell fraction numerator 6 straight a over denominator begin display style 3 over 5 end style end fraction end cell equals 10 row cell 6 straight a cross times 5 end cell equals cell 10 cross times 3 end cell row cell 30 straight a end cell equals 30 row straight a equals cell 30 over 30 end cell row straight a equals 1 end table end style 

Jadi, nilai begin mathsize 14px style straight a end style adalah 1

Jadi, jawaban yang tepat adalah A

0

Roboguru

Nilai x→5lim​x−5x​−5​​ adalah ...

Pembahasan Soal:

Terlebih dahulu kita substitusi nilai x = 5 pada limit tersebut, diperoleh:

limit as x rightwards arrow 5 of fraction numerator square root of x minus square root of 5 over denominator x minus 5 end fraction equals fraction numerator square root of 5 minus square root of 5 over denominator 5 minus 5 end fraction equals 0 over 0 

Karena limit tersebut bentuk tak tentu maka tipe soal limit seperti ini, dapat menggunakan perkalian akar sekawan karena pada soal terdapat operasi yang melibatkan akar, sehingga:

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 square root of x minus square root of 5 over denominator x minus 5 end fraction cross times fraction numerator square root of x plus square root of 5 over denominator square root of x plus square root of 5 end fraction end cell row blank equals cell limit as x rightwards arrow 5 of fraction numerator up diagonal strike left parenthesis x minus 5 right parenthesis end strike over denominator up diagonal strike left parenthesis x minus 5 right parenthesis end strike left parenthesis square root of x plus square root of 5 right parenthesis end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 5 plus square root of 5 end fraction end cell row blank equals cell fraction numerator 1 over denominator 2 square root of 5 end fraction cross times fraction numerator square root of 5 over denominator square root of 5 end fraction end cell row blank equals cell fraction numerator square root of 5 over denominator 10 end fraction end cell end table end style

Dengan demikian, hasil dari limit as x rightwards arrow 5 of fraction numerator square root of x minus square root of 5 over denominator x minus 5 end fraction equals fraction numerator square root of 5 over denominator 10 end fraction.

Oleh karena itu, jawaban yang benar adalah C 

0

Roboguru

Hitunglah nilai dari x→2lim​2x+5​−5x−1​x−2​.

Pembahasan Soal:

Dengan metode substitusi diperoleh sebagai berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator x minus 2 over denominator square root of 2 x plus 5 end root minus square root of 5 x minus 1 end root end fraction end cell equals cell fraction numerator 2 minus 2 over denominator square root of 2 times 2 plus 5 end root minus square root of 5 times 2 minus 1 end root end fraction end cell row blank equals cell 0 over 0 end cell end table end style

Berdasarkan uraian di atas, diperoleh hasilnya tak tentu atau 0 over 0, maka nilai limit tersebut harus ditentukan dengan metode yang lain. Dengan metode mengalikan akar sekawan terlebih dahulu lalu substitusi, diperoleh sebagai berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 2 of fraction numerator x minus 2 over denominator square root of 2 x plus 5 end root minus square root of 5 x minus 1 end root end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator x minus 2 over denominator square root of 2 x plus 5 end root minus square root of 5 x minus 1 end root end fraction times fraction numerator square root of 2 x plus 5 end root plus square root of 5 x minus 1 end root over denominator square root of 2 x plus 5 end root plus square root of 5 x minus 1 end root end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis square root of 2 x plus 5 end root plus square root of 5 x minus 1 end root right parenthesis over denominator left parenthesis 2 x plus 5 right parenthesis minus left parenthesis 5 x minus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis square root of 2 x plus 5 end root plus square root of 5 x minus 1 end root right parenthesis over denominator 2 x plus 5 minus 5 x plus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis square root of 2 x plus 5 end root plus square root of 5 x minus 1 end root right parenthesis over denominator negative 3 x plus 6 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis square root of 2 x plus 5 end root plus square root of 5 x minus 1 end root right parenthesis over denominator negative 3 left parenthesis x minus 2 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator square root of 2 x plus 5 end root plus square root of 5 x minus 1 end root over denominator negative 3 end fraction end cell row blank equals cell fraction numerator square root of 2 times 2 plus 5 end root plus square root of 5 times 2 minus 1 end root over denominator negative 3 end fraction end cell row blank equals cell fraction numerator 3 plus 3 over denominator negative 3 end fraction end cell row blank equals cell fraction numerator 6 over denominator negative 3 end fraction end cell row blank equals cell negative 2 end cell 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 x minus 2 over denominator square root of 2 x plus 5 end root minus square root of 5 x minus 1 end root 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 minus end table 2 end style.

0

Roboguru

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