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Pertanyaan

Nilai ​ ​ lim x → 4 ​ ​ ​ ​ 1 − x − 3 ​ x 2 − 16 ​ ​ adalah ....

Nilai  adalah ....

  1. begin mathsize 14px style negative 16 end style 

  2.  size 14px minus size 14px 4 

  3. 4 

  4. begin mathsize 14px style 16 end style 

  5. begin mathsize 14px style 32 end style 

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D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

Jawaban yang benar adalah A

Jawaban yang benar adalah A

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Pembahasan

Pembahasan
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Dengan menerapkan konsep perkalian sekawan; Jadi, Nilai dari . Oleh karena itu, Jawaban yang benar adalah A

Dengan menerapkan konsep perkalian sekawan;

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 4 of fraction numerator straight x squared minus 16 over denominator 1 minus square root of straight x minus 3 end root end fraction end cell equals cell limit as straight x rightwards arrow 4 of fraction numerator left parenthesis straight x squared minus 16 right parenthesis over denominator 1 minus square root of straight x minus 3 end root end fraction times fraction numerator left parenthesis 1 plus square root of straight x minus 3 end root right parenthesis over denominator left parenthesis 1 plus square root of straight x minus 3 end root right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 4 of fraction numerator left parenthesis straight x squared minus 16 right parenthesis left parenthesis 1 plus square root of straight x minus 3 end root right parenthesis over denominator left parenthesis 1 squared minus left parenthesis square root of straight x minus 3 end root right parenthesis squared end fraction end cell row blank equals cell limit as straight x rightwards arrow 4 of fraction numerator left parenthesis straight x squared minus 16 right parenthesis left parenthesis 1 plus square root of straight x minus 3 end root right parenthesis over denominator 1 minus left parenthesis straight x minus 3 right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 4 of fraction numerator left parenthesis straight x squared minus 16 right parenthesis left parenthesis 1 plus square root of straight x minus 3 end root right parenthesis over denominator 1 minus straight x plus 3 end fraction end cell row blank equals cell limit as straight x rightwards arrow 4 of fraction numerator left parenthesis straight x squared minus 16 right parenthesis left parenthesis 1 plus square root of straight x minus 3 end root right parenthesis over denominator 4 minus straight x end fraction end cell row blank equals cell limit as straight x rightwards arrow 4 of fraction numerator left parenthesis straight x minus 4 right parenthesis left parenthesis straight x plus 4 right parenthesis left parenthesis 1 plus square root of straight x minus 3 end root right parenthesis over denominator negative left parenthesis straight x minus 4 right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 4 of fraction numerator left parenthesis straight x plus 4 right parenthesis left parenthesis 1 plus square root of straight x minus 3 end root right parenthesis over denominator negative 1 end fraction end cell row blank equals cell fraction numerator left parenthesis 4 plus 4 right parenthesis left parenthesis 1 plus square root of 4 minus 3 end root right parenthesis over denominator negative 1 end fraction end cell row blank equals cell fraction numerator 8 times 2 over denominator negative 1 end fraction end cell row blank equals cell negative 16 end cell end table end style  

Jadi, Nilai dari undefined.

Oleh karena itu, Jawaban yang benar adalah A

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Nayla Nazwa

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