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Nilai

Nilai begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator square root of 4 plus 2 x end root minus square root of 4 minus 2 x end root over denominator x end fraction equals... end style 

  1. begin mathsize 14px style 4 end style 

  2. begin mathsize 14px style 2 end style 

  3. begin mathsize 14px style 1 end style 

  4. begin mathsize 14px style 0 end style 

  5. begin mathsize 14px style negative 1 end style 

Jawaban:

Pada soal limit tersebut terdapat bentuk akar, sehingga untuk menentukan nilai limit tersebut dengan mengalikan dengan akar sekawan.

Apabila begin mathsize 14px style x equals 0 end style akan diperoleh bentuk begin mathsize 14px style 0 over 0 end style, atau bentuk tak tentu suatu limit, sehingga bentuk limit tersebut harus dikalikan dengan akar sekawannya, seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row blank blank cell limit as x rightwards arrow 0 of fraction numerator square root of 4 plus 2 x end root minus square root of 4 minus 2 x end root over denominator x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator square root of 4 plus 2 x end root minus square root of 4 minus 2 x end root over denominator x end fraction cross times fraction numerator square root of 4 plus 2 x end root plus square root of 4 minus 2 x end root over denominator square root of 4 plus 2 x end root plus square root of 4 minus 2 x end root end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator open parentheses 4 plus 2 x close parentheses minus open parentheses 4 minus 2 x close parentheses over denominator x open parentheses square root of 4 plus 2 x end root plus square root of 4 minus 2 x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 4 plus 2 x minus 4 plus 2 x over denominator x open parentheses square root of 4 plus 2 x end root plus square root of 4 minus 2 x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 4 x over denominator x open parentheses square root of 4 plus 2 x end root plus square root of 4 minus 2 x end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 4 over denominator open parentheses square root of 4 plus 2 x end root plus square root of 4 minus 2 x end root close parentheses end fraction end cell row blank equals cell fraction numerator 4 over denominator open parentheses square root of 4 plus 2 open parentheses 0 close parentheses end root plus square root of 4 minus 2 open parentheses 0 close parentheses end root close parentheses end fraction end cell row blank equals cell fraction numerator 4 over denominator square root of 4 plus square root of 4 end fraction end cell row blank equals cell fraction numerator 4 over denominator 2 plus 2 end fraction end cell row blank equals cell 4 over 4 end cell row blank equals 1 end table end style 

Sehingga diperoleh begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator square root of 4 plus 2 x end root minus square root of 4 minus 2 x end root over denominator x end fraction equals 1 end style.

Jadi, jawaban yang tepat adalah C. 

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