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Pertanyaan

limit as x rightwards arrow infinity of space open parentheses 4 square root of x squared plus x minus 3 end root minus 4 x plus 2 close parentheses equals horizontal ellipsis     

  1. negative infinity

  2. negative 4 

  3. 4 

  4. 8 

  5. infinity 

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Perhatikan bentuk limit tak hingga berikut :

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 L

L merupakan hasil penyelesaian limit, maka

jika a greater than p maka L equals infinity,
jika a equals p maka L equals fraction numerator b minus q over denominator 2 square root of a end fraction, dan
jika a less than p maka L equals negative infinity.

Pada soal diberikan limit as x rightwards arrow infinity of space open parentheses 4 square root of x squared plus x minus 3 end root minus 4 x plus 2 close parentheses. Ubah fungsinya seperti bentuk umum limit di atas, maka

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of space open parentheses 4 square root of x squared plus x minus 3 end root minus 4 x plus 2 close parentheses end cell row blank equals cell limit as x rightwards arrow infinity of space open parentheses square root of 4 squared open parentheses x squared plus x minus 3 close parentheses end root minus left parenthesis 4 x minus 2 right parenthesis close parentheses end cell row blank equals cell limit as x rightwards arrow infinity of space open parentheses square root of 4 squared open parentheses x squared plus x minus 3 close parentheses end root minus square root of left parenthesis 4 x minus 2 right parenthesis squared end root close parentheses end cell row blank equals cell limit as x rightwards arrow infinity of space open parentheses square root of 16 open parentheses x squared plus x minus 3 close parentheses end root minus square root of 16 x squared minus 16 x plus 4 end root close parentheses end cell row blank equals cell limit as x rightwards arrow infinity of space open parentheses square root of 16 x squared plus 16 x minus 48 end root minus square root of 16 x squared minus 16 x plus 4 end root close parentheses end cell end table end style

Berdasarkan bentuk umum limit, a equals p equals 16. Sedangkan nilai b equals 16 dan q equals negative 16,sehingga

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of space open parentheses square root of 16 x squared plus 16 x minus 48 end root minus square root of 16 x squared minus 16 x plus 4 end root close parentheses end cell row blank equals cell fraction numerator b minus q over denominator 2 square root of a end fraction end cell row blank equals cell fraction numerator 16 minus left parenthesis negative 16 right parenthesis over denominator 2 square root of 16 end fraction end cell row blank equals cell fraction numerator 16 plus 16 over denominator 2 times 4 end fraction end cell row blank equals cell 32 over 8 end cell row blank equals 4 end table end style

Dengan demikian, hasil dari limit tak hingga tersebut adalah 4.

Jadi, jawaban yang tepat adalah C.

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