Pertanyaan

x → 1 lim ​ 1 − x 2 x 2 + 3 ​ − x − 1 ​ = ...

 

  1. begin mathsize 14px style negative 1 half end style 

  2. begin mathsize 14px style negative 1 fourth end style 

  3. undefined 

  4. begin mathsize 14px style 1 fourth end style 

  5. begin mathsize 14px style 1 half end style 

A. Damanhuri

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

jawaban yang tepat adalah D.

Pembahasan

Pembahasan

Jika dijumpai limit dengan bentuk akar dan tidak dapat difaktorkan secara langsung, maka kalikan dengan bentuk sekawannya. Dengan demikian, . Jadi, jawaban yang tepat adalah D.

Jika dijumpai limit dengan bentuk akar dan tidak dapat difaktorkan secara langsung, maka kalikan dengan bentuk sekawannya.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 1 of invisible function application fraction numerator square root of x squared plus 3 end root minus x minus 1 over denominator 1 minus x squared end fraction end cell row blank equals cell limit as x rightwards arrow 1 of invisible function application fraction numerator square root of x squared plus 3 end root minus left parenthesis x plus 1 right parenthesis over denominator 1 minus x squared end fraction times fraction numerator square root of x squared plus 3 end root plus left parenthesis x plus 1 right parenthesis over denominator square root of x squared plus 3 end root plus left parenthesis x plus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator x squared plus 3 minus left parenthesis x plus 1 right parenthesis squared over denominator open parentheses 1 minus x squared close parentheses open parentheses square root of x squared plus 3 end root plus left parenthesis x plus 1 right parenthesis close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator x squared plus 3 minus left parenthesis x squared plus 2 x plus 1 right parenthesis over denominator open parentheses 1 minus x squared close parentheses open parentheses square root of x squared plus 3 end root plus left parenthesis x plus 1 right parenthesis close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator negative 2 x plus 2 over denominator open parentheses 1 minus x close parentheses left parenthesis 1 plus x right parenthesis open parentheses square root of x squared plus 3 end root plus left parenthesis x plus 1 right parenthesis close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of space fraction numerator negative 2 up diagonal strike left parenthesis x minus 1 right parenthesis end strike over denominator negative up diagonal strike left parenthesis x minus 1 right parenthesis end strike left parenthesis 1 plus x right parenthesis open parentheses square root of x squared plus 3 end root plus left parenthesis x plus 1 right parenthesis close parentheses end fraction end cell row blank equals cell fraction numerator negative 2 over denominator negative left parenthesis 1 plus 1 right parenthesis open parentheses square root of 1 squared plus 3 end root plus left parenthesis 1 plus 1 right parenthesis close parentheses end fraction end cell row blank equals cell fraction numerator up diagonal strike negative 2 end strike over denominator up diagonal strike negative 2 end strike left parenthesis 2 plus 2 right parenthesis end fraction end cell row blank equals cell 1 fourth end cell end table end style 

Dengan demikian, begin mathsize 14px style limit as x rightwards arrow 1 of invisible function application fraction numerator square root of x squared plus 3 end root minus x minus 1 over denominator 1 minus x squared end fraction equals 1 fourth end style.

Jadi, jawaban yang tepat adalah D.

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