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Nilai dari: begin mathsize 14px style limit as straight x rightwards arrow infinity of space square root of 4 straight x squared minus 12 straight x minus 1 end root minus 2 straight x plus 1 equals.... end style  

Pembahasan Video:

Pembahasan Soal:

Konsep limit  tak hingga: metode membagi dengan pangkat tertinggi

Diketahui: begin mathsize 12px style limit as straight x rightwards arrow infinity of space square root of 4 straight x squared minus 12 straight x minus 1 end root minus 2 straight x plus 1 equals.... end style 

Fungsi di atas belum fungsi bentuk rasional,, dengan perkalian akar sekawan maka:

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

Pangkat tertinggi adalah begin mathsize 14px style square root of straight x squared end root end style atau sama dengan begin mathsize 14px style x end style, maka pembilang dan penyebut dibagi oleh begin mathsize 14px style x end style, maka diperoleh penyelesaiannya sebagai berikut:

   begin mathsize 11px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow infinity of fraction numerator negative 8 straight x minus 2 over denominator square root of 4 straight x squared minus 12 straight x minus 1 end root plus left parenthesis 2 straight x minus 1 right parenthesis end fraction end cell equals cell limit as straight x rightwards arrow infinity of fraction numerator begin display style fraction numerator negative 8 straight x over denominator straight x end fraction minus 2 over straight x end style over denominator square root of begin display style fraction numerator 4 straight x squared over denominator straight x squared end fraction minus fraction numerator 12 straight x over denominator straight x squared end fraction minus 1 over straight x squared end style end root plus left parenthesis begin display style fraction numerator 2 straight x over denominator straight x end fraction end style minus begin display style 1 over straight x end style right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow infinity of fraction numerator begin display style negative 8 minus 2 over straight x end style over denominator square root of begin display style 4 minus 12 over straight x minus 1 over straight x squared end style end root plus left parenthesis begin display style 2 end style minus begin display style 1 over straight x end style right parenthesis end fraction end cell row blank equals cell fraction numerator begin display style negative 8 minus 2 over infinity end style over denominator square root of begin display style 4 minus 12 over infinity minus 1 over infinity squared end style end root plus left parenthesis begin display style 2 end style minus begin display style 1 over infinity end style right parenthesis end fraction end cell row blank equals cell fraction numerator begin display style negative 8 minus 0 end style over denominator square root of begin display style 4 minus 0 minus 0 end style end root plus left parenthesis begin display style 2 end style minus begin display style 0 end style right parenthesis end fraction end cell row blank equals cell fraction numerator negative 8 over denominator 2 plus 2 end fraction end cell row blank equals cell negative 2 end cell end table end style   

Jadi, Nilai  begin mathsize 14px style limit as straight x rightwards arrow infinity of space square root of 4 straight x squared minus 12 straight x minus 1 end root minus 2 straight x plus 1 equals negative 2 end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 14 September 2021

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Pertanyaan yang serupa

Nilai dari  adalah...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of a x squared plus b end root minus square root of a x squared plus q end root end cell equals cell fraction numerator b minus q over denominator 2 square root of a end fraction end cell row cell limit as x rightwards arrow infinity of open parentheses x minus square root of x squared minus 4 x end root close parentheses end cell equals cell square root of x squared end root minus square root of x squared minus 4 x end root end cell row a equals 1 row b equals 0 row q equals cell negative 4 end cell row cell fraction numerator b minus q over denominator 2 square root of a end fraction end cell equals cell fraction numerator 0 minus left parenthesis negative 4 right parenthesis over denominator 2 square root of 1 end fraction end cell row blank equals cell 4 over 2 end cell row blank equals 2 row blank blank blank end table

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Hasil dari   jika a < p adalah ...

Pembahasan Soal:

Perhatikan bahwa hasil darisize 14px lim with size 14px x size 14px rightwards arrow size 14px infinity below size 14px invisible function application open parentheses square root of size 14px a size 14px x to the power of size 14px 2 size 14px plus size 14px b size 14px x size 14px plus size 14px c end root size 14px minus square root of size 14px p size 14px x to the power of size 14px 2 size 14px plus size 14px q size 14px x size 14px plus size 14px r end root close parentheses    bergantung terhadap hubungan a  dan p  sebagai berikut.

Jika a > p, maka size 14px lim with size 14px x size 14px rightwards arrow size 14px infinity below size 14px invisible function application open parentheses square root of size 14px a size 14px x to the power of size 14px 2 size 14px plus size 14px b size 14px x size 14px plus size 14px c end root size 14px minus square root of size 14px p size 14px x to the power of size 14px 2 size 14px plus size 14px q size 14px x size 14px plus size 14px r end root close parentheses size 14px equals size 14px infinity 

Jika a < p, maka size 14px lim with size 14px x size 14px rightwards arrow size 14px infinity below size 14px invisible function application open parentheses square root of size 14px a size 14px x to the power of size 14px 2 size 14px plus size 14px b size 14px x size 14px plus size 14px c end root size 14px minus square root of size 14px p size 14px x to the power of size 14px 2 size 14px plus size 14px q size 14px x size 14px plus size 14px r end root close parentheses size 14px equals size 14px minus size 14px infinity  

Dan jika a = p, maka size 14px lim with size 14px x size 14px rightwards arrow size 14px infinity below size 14px invisible function application open parentheses square root of size 14px a size 14px x to the power of size 14px 2 size 14px plus size 14px b size 14px x size 14px plus size 14px c end root size 14px minus square root of size 14px p size 14px x to the power of size 14px 2 size 14px plus size 14px q size 14px x size 14px plus size 14px r end root close parentheses size 14px equals fraction numerator size 14px b size 14px minus size 14px q over denominator size 14px 2 square root of size 14px a end fraction  

Sehingga, karena a < p, maka

undefined   

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Pembahasan Soal:

Cara menghitung limit tak hingga pada fungsi rasional yang memuat polinomial salah satunya adalah membagi setiap suku pada pembilang dan penyebut dengan pangkat tertinggi dari penyebut sebagai berikut:

ingat bahwa square root of x to the power of 4 end root equals x squared

limx16x4+1(4x+2)2======limx16x4+116x2+16x+4×x21x21limxx416x4+x41x216x2+x216x+x24limx16+x4116+x16+x2416+016+0+04164

Dengan demikian hasilnya adalah begin mathsize 14px style 4 end style.

 

1

Roboguru

Hasil  adalah ...

Pembahasan Soal:

Terdapat beberapa cara untuk menentukan limit menuju tak hingga dengan bentuk akar, seperti berikut ini:

begin mathsize 14px 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 infinity row cell fraction numerator b minus q over denominator 2 square root of a end fraction end cell row cell negative infinity end cell end table close end style 

Diperoleh begin mathsize 14px style infinity end style apabila begin mathsize 14px style a greater than p end style, diperoleh begin mathsize 14px style fraction numerator b minus q over denominator 2 square root of a end fraction end style untuk begin mathsize 14px style a equals p end style, dan begin mathsize 14px style negative infinity end style untuk begin mathsize 14px style a less than p end style

begin mathsize 14px style limit as x rightwards arrow infinity of open parentheses square root of x squared minus 2 x plus 3 end root minus square root of x squared plus 4 x minus 1 end root close parentheses end style, mempunyai bentuk selisih akar kuadrat seperti bentuk di atas, dimana begin mathsize 14px style a equals p equals 1 end stylebegin mathsize 14px style b equals negative 2 end stylebegin mathsize 14px style q equals 4 end style. Oleh karena begin mathsize 14px style a equals p end style maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator b minus q over denominator 2 square root of a end fraction end cell equals cell fraction numerator negative 2 minus 4 over denominator 2 square root of 1 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 

Sehingga hasil begin mathsize 14px style limit as x rightwards arrow infinity of left parenthesis square root of x squared minus 2 x plus 3 end root minus square root of x squared plus 4 x minus 1 end root right parenthesis equals negative 3 end style.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

= ....

Pembahasan Soal:

limit as x rightwards arrow infinity of open parentheses square root of 2 x squared plus 3 x minus 1 end root minus 2 x squared minus 5 x space plus 3 close parentheses equals fraction numerator negative 3 minus left parenthesis negative 5 right parenthesis over denominator 2 square root of 2 end fraction equals fraction numerator 8 over denominator square root of 2 end fraction equals 2 square root of 2

0

Roboguru

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