Roboguru

Tentukan nilai:

Pertanyaan

Tentukan nilai:

begin mathsize 14px style limit as straight x rightwards arrow infinity of square root of straight x left parenthesis square root of straight x minus 4 end root minus square root of straight x plus 2 end root right parenthesis end style 

Pembahasan Soal:

Dengan sifat distributif, maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow infinity of square root of straight x left parenthesis square root of straight x minus 4 end root minus square root of straight x plus 2 end root right parenthesis end cell equals cell limit as straight x rightwards arrow infinity of left parenthesis square root of straight x left parenthesis straight x minus 4 right parenthesis end root minus square root of straight x left parenthesis straight x plus 2 right parenthesis end root right parenthesis end cell row blank equals cell limit as straight x rightwards arrow infinity of left parenthesis square root of straight x squared minus 4 x end root minus square root of straight x squared plus 2 x end root end cell row blank blank blank end table end style 

Dengan metode perkalian sekawan:

begin mathsize 14px style limit as straight x rightwards arrow infinity of left parenthesis square root of straight x squared minus 4 straight x end root minus square root of straight x squared plus 2 straight x end root equals limit as straight x rightwards arrow infinity of left parenthesis square root of straight x squared minus 4 straight x end root minus square root of straight x squared plus 2 straight x end root right parenthesis times fraction numerator left parenthesis square root of straight x squared minus 4 straight x end root plus square root of straight x squared plus 2 straight x end root right parenthesis over denominator left parenthesis square root of straight x squared minus 4 straight x end root plus square root of straight x squared plus 2 straight x end root right parenthesis end fraction equals limit as straight x rightwards arrow infinity of fraction numerator left parenthesis straight x squared minus 4 straight x right parenthesis minus left parenthesis straight x squared plus 2 straight x right parenthesis over denominator left parenthesis square root of straight x squared minus 4 straight x end root plus square root of straight x squared plus 2 straight x end root end fraction equals limit as straight x rightwards arrow infinity of fraction numerator negative 6 straight x over denominator left parenthesis square root of straight x squared minus 4 straight x end root plus square root of straight x squared plus 2 straight x end root end fraction end style 

variabel undefined dengan pangkat terbesar dari begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator negative 6 straight x over denominator left parenthesis square root of straight x squared minus 4 straight x end root plus square root of straight x squared plus 2 straight x end root end fraction end cell end table end style adalah undefined maka pembilang dan penyebut pada limit fungsi dibagi oleh undefined. sehingga diperoleh:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow infinity of fraction numerator negative 6 straight x over denominator left parenthesis square root of straight x squared minus 4 straight x end root plus square root of straight x squared plus 2 straight x end root end fraction end cell equals cell limit as straight x rightwards arrow infinity of fraction numerator begin display style fraction numerator negative 6 straight x over denominator straight x end fraction end style over denominator left parenthesis begin display style fraction numerator square root of straight x squared minus 4 straight x end root over denominator straight x end fraction end style plus begin display style fraction numerator square root of straight x squared plus 2 straight x end root over denominator straight x end fraction end style end fraction end cell row blank equals cell limit as straight x rightwards arrow infinity of fraction numerator begin display style fraction numerator negative 6 straight x over denominator straight x end fraction end style over denominator left parenthesis begin display style fraction numerator square root of straight x squared minus 4 straight x end root over denominator square root of straight x squared end root end fraction end style plus begin display style fraction numerator square root of straight x squared plus 2 straight x end root over denominator square root of straight x squared end root end fraction end style end fraction end cell row blank blank cell limit as straight x rightwards arrow infinity of fraction numerator begin display style negative 6 end style over denominator begin display style square root of fraction numerator begin display style straight x squared minus 4 straight x end style over denominator begin display style straight x squared end style end fraction end root plus square root of fraction numerator begin display style straight x squared plus 2 straight x end style over denominator begin display style straight x squared end style end fraction end root end style end fraction end cell row blank equals cell limit as straight x rightwards arrow infinity of fraction numerator begin display style negative 6 end style over denominator begin display style square root of 1 minus fraction numerator begin display style 4 end style over denominator begin display style straight x end style end fraction end root plus square root of 1 plus fraction numerator begin display style 2 end style over denominator begin display style straight x end style end fraction end root end style end fraction end cell row blank equals cell fraction numerator begin display style negative 6 end style over denominator begin display style square root of 1 minus fraction numerator begin display style 4 end style over denominator begin display style infinity end style end fraction end root plus square root of 1 plus fraction numerator begin display style 2 end style over denominator begin display style infinity end style end fraction end root end style end fraction end cell row blank equals cell fraction numerator begin display style negative 6 end style over denominator begin display style square root of 1 minus 0 end root plus square root of 1 plus 0 end root end style end fraction end cell row blank equals cell fraction numerator negative 6 over denominator 2 end fraction end cell row blank equals cell negative 3 end cell end table end style 

Jadi, Nilai Error converting from MathML to accessible text..

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. RGFLLIMA

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

Tentukan nilai dari  ....

Pembahasan Soal:

Ingat kembali rumus penyelesaian limit tak hingga berikut.

begin mathsize 12px style limit as x rightwards arrow infinity of open parentheses square root of a x squared plus b x plus c end root minus square root of p x squared plus q x plus r end root close parentheses equals open curly brackets table attributes columnalign left end attributes row cell infinity comma space untuk space a greater than p end cell row cell fraction numerator b minus q over denominator 2 square root of a end fraction comma space untuk space a equals p end cell row cell negative infinity comma space untuk space a less than p end cell end table close end style 

Terlebih dahulu kita jabarkan soal tersebut sehingga memiliki bentuk seperti persamaan di atas, yaitu sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of square root of x squared minus x end root minus x plus 3 end cell equals cell limit as x rightwards arrow infinity of square root of x squared minus x end root minus square root of open parentheses x plus 3 close parentheses open parentheses x plus 3 close parentheses end root end cell row blank equals cell limit as x rightwards arrow infinity of square root of x squared minus x end root minus square root of x squared plus 6 x plus 9 end root end cell end table 

Karena nilai a = p = 1 maka berlaku aturan kedua, sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of square root of x squared minus x end root minus x plus 3 end cell equals cell limit as x rightwards arrow infinity of square root of x squared minus x end root minus square root of x squared plus 6 x plus 9 end root end cell row blank equals cell fraction numerator b minus q over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative 1 minus 6 over denominator 2 cross times 1 end fraction end cell row blank equals cell fraction numerator negative 7 over denominator 2 end fraction end cell end table  

Dengan demikian, nilai dari limit as straight x rightwards arrow infinity of square root of x squared minus x end root minus x plus 3 equals negative 7 over 2.

Roboguru

Nilai

Pembahasan Soal:

Langkah-langkah limit tak hingga bentuk akar:

  1. Kali dengan akar sekawan.
  2. Bagi dengan suku berpangkat tertinggi dari pembilang atau penyebut.
  3. Substitusi x mendekati infinity.

Langkah pertama (kali dengan akar sekawan):

Ingat: left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis equals a squared minus b squared.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of open parentheses x minus square root of x squared minus 5 x end root close parentheses end cell equals cell limit as x rightwards arrow infinity of open parentheses x minus square root of x squared minus 5 x end root close parentheses cross times fraction numerator open parentheses x plus square root of x squared minus 5 x end root close parentheses over denominator open parentheses x plus square root of x squared minus 5 x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator open parentheses x squared minus open parentheses square root of x squared minus 5 x end root close parentheses squared close parentheses over denominator open parentheses x plus square root of x squared minus 5 x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator open parentheses x squared minus x squared plus 5 x close parentheses over denominator open parentheses x plus square root of x squared minus 5 x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 5 x over denominator x plus square root of x squared minus 5 x end root end fraction end cell end table

Langkah kedua (bagi dengan suku berpangkat tertinggi dari pembilang atau penyebut):

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

Langkah ketiga (substitusi x mendekati infinity):

Ingat: limit as x rightwards arrow infinity of k over x to the power of n equals k over infinity to the power of n equals 0.

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

Jadi table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses x minus square root of x squared minus 5 x end root close parentheses end cell end table equals 2 comma 5.

Oleh karena itu, jawaban yang tepat adalah D.

Roboguru

9.

Pembahasan Soal:

Penyelesaikan limit dengan cara berikut.

begin mathsize 14px style limit as x rightwards arrow infinity of square root of a x squared plus b c plus c end root minus square root of p x squared plus q x plus r end root equals open curly brackets table attributes columnalign left end attributes row cell plus infinity space text untuk end text space a greater than p end cell row cell negative infinity text  untuk end text space a less than p end cell row cell fraction numerator b minus q over denominator 2 square root of a end fraction space text untuk end text space a equals p end cell end table close end style 

Untuk begin mathsize 14px style limit as x rightwards arrow infinity of open parentheses 2 x minus 2 minus square root of 4 x squared minus 4 x plus 3 end root close parentheses end style, diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of open parentheses 2 x minus 2 minus square root of 4 x squared minus 4 x plus 3 end root close parentheses end cell equals cell limit as x rightwards arrow infinity of square root of open parentheses 2 x minus 2 close parentheses squared end root minus square root of 4 x squared minus 4 x plus 3 end root end cell row blank equals cell limit as x rightwards arrow infinity of square root of 4 x squared minus 8 x plus 4 end root minus square root of 4 x squared minus 4 x plus 3 end root end cell row blank equals cell fraction numerator negative 8 minus open parentheses negative 4 close parentheses over denominator 2 square root of 4 end fraction end cell row blank equals cell fraction numerator negative 4 over denominator 4 end fraction end cell row blank equals cell negative 1 end cell end table end style 

Jadi, nilai begin mathsize 14px style limit as x rightwards arrow infinity of open parentheses 2 x minus 2 minus square root of 4 x squared minus 4 x plus 3 end root close parentheses end style adalah begin mathsize 14px style negative 1 end style.

Roboguru

Tentukan nilai limit berikut.

Pembahasan Soal:

Mencari nilai limit:

limit as x rightwards arrow infinity thin space of open parentheses fraction numerator 7 x over denominator 3 minus square root of 9 plus 1 half x end root end fraction close parentheses equals 7 times limit as x rightwards arrow infinity thin space of open parentheses fraction numerator x over denominator 3 minus square root of 9 plus 1 half x end root end fraction times fraction numerator 3 plus square root of 9 plus 1 half x end root over denominator 3 plus square root of 9 plus 1 half x end root end fraction close parentheses equals 7 times limit as x rightwards arrow infinity thin space of open parentheses fraction numerator x open parentheses 3 plus square root of 9 plus 1 half x end root close parentheses over denominator 9 minus 9 minus 1 half x end fraction close parentheses equals 7 times limit as x rightwards arrow infinity thin space of open parentheses fraction numerator x open parentheses 3 plus square root of 9 plus 1 half x end root close parentheses over denominator negative 1 half x end fraction close parentheses equals 7 times limit as x rightwards arrow infinity thin space of open parentheses negative 2 open parentheses 3 plus square root of 9 plus 1 half x end root close parentheses close parentheses equals equals 7 times limit as x rightwards arrow infinity thin space of open parentheses negative 2 open parentheses 3 plus square root of 1 half open parentheses x plus 18 close parentheses end root close parentheses close parentheses equals 7 times open parentheses negative 2 close parentheses open parentheses 3 plus square root of 1 half open parentheses x plus 18 close parentheses end root close parentheses equals 7 open parentheses negative 2 close parentheses open parentheses 3 plus square root of 1 half open parentheses infinity close parentheses end root close parentheses equals 7 open parentheses negative 2 close parentheses left parenthesis 3 plus infinity right parenthesis equals 7 open parentheses negative 2 right parenthesis times left parenthesis infinity thin space close parentheses equals negative infinity thin space  

Jadi, limit as x rightwards arrow infinity thin space of open parentheses fraction numerator 7 x over denominator 3 minus square root of 9 plus 1 half x end root end fraction close parentheses equals negative infinity thin space.

Roboguru

Nilai dari  adalah ....

Pembahasan Soal:

Misalkan 1 over x equals p 

Sehingga mengakibatkan batas x juga berubah menjadi  p rightwards arrow infinity.

Diperoleh persamaan baru yaitu:

limit as p rightwards arrow infinity of square root of 4 plus 8 p plus p squared end root minus square root of 4 minus 4 p plus p squared end root 

Untuk bentuk limit diatas berlaku rumus berikut:

limit as x rightwards arrow infinity of square root of a x squared plus b x plus c end root minus square root of p x squared plus q x plus r end root 

table attributes columnalign right center left columnspacing 0px end attributes row L equals cell negative infinity space jika space dan space hanya space jika space a less than p end cell row L equals cell fraction numerator b minus q over denominator 2 square root of a end fraction space jika space dan space hanya space jika space a equals p end cell row L equals cell infinity space jika space dan space hanya space jika space a greater than p end cell end table 

Dari persamaan soal yang baru diketahui bahwa a = p = 1, maka berlaku rumus kedua.

 table attributes columnalign right center left columnspacing 0px end attributes row L equals cell fraction numerator 8 minus open parentheses negative 4 close parentheses over denominator 2 square root of 1 end fraction end cell row blank equals cell fraction numerator 8 plus 4 over denominator 2 end fraction end cell row blank equals cell 12 over 2 end cell row blank equals 6 end table 

Dengan demikian, limit as p rightwards arrow infinity of square root of 4 plus 8 p plus p squared end root minus square root of 4 minus 4 p plus p squared end root equals 6.

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