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lim x → 1 ​ 1 − 3 x ​ 1 − x ​ ​ = ...

 ...

  1. begin mathsize 14px style 3 over 4 end style 

  2. begin mathsize 14px style 2 over 3 end style 

  3. begin mathsize 14px style 1 half end style 

  4. begin mathsize 14px style 3 over 2 end style 

  5. begin mathsize 14px style 4 over 3 end style 

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R. RGFLSATU

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jawabannya adalah D.

jawabannya adalah D.

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Pembahasan

Perhatikan bahwa jika dilakukan substitusi , maka didapatkan hasil . Sehingga perlu dicari cara lain untuk mencari nilai limitnya. Misalkan . Maka untuk maka didapat bahwa . Sehingga Perhatikan bahwa . Karena .Maka Perhatikan bahwa untuk , sebenarnya didapat pula bahwa . Sehingga Perhatikan bahwa . Karena .Maka Maka Jadi, jawabannya adalah D.

Perhatikan bahwa jika dilakukan substitusi begin mathsize 14px style x equals 1 end style, maka didapatkan hasil begin mathsize 14px style 0 over 0 end style .
Sehingga perlu dicari cara lain untuk mencari nilai limitnya.

Misalkan begin mathsize 14px style straight x equals straight y to the power of 6 end style .
Maka untuk begin mathsize 14px style straight x rightwards arrow 1 end style   maka didapat bahwa begin mathsize 14px style straight y rightwards arrow 1 end style  .
Sehingga

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals limit as straight y rightwards arrow 1 of fraction numerator 1 minus square root of straight y to the power of 6 end root over denominator 1 minus cube root of straight y to the power of 6 end root end fraction end style 

Perhatikan bahwa begin mathsize 14px style square root of straight y to the power of 6 end root equals open vertical bar straight y cubed close vertical bar end style . Karena begin mathsize 14px style straight y rightwards arrow 1 comma space maka space straight y greater than 0 comma space sehingga square root of straight y to the power of 6 end root equals straight y cubed end style .Maka

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals limit as straight y rightwards arrow 1 of fraction numerator 1 minus square root of straight y to the power of 6 end root over denominator 1 minus cube root of straight y to the power of 6 end root end fraction equals limit as straight y rightwards arrow 1 of fraction numerator 1 minus straight y cubed over denominator 1 minus straight y squared end fraction equals limit as straight y rightwards arrow 1 of fraction numerator open parentheses 1 minus straight y close parentheses open parentheses 1 plus straight y plus straight y squared close parentheses over denominator open parentheses 1 plus straight y close parentheses open parentheses 1 minus straight y close parentheses end fraction equals limit as straight y rightwards arrow 1 of fraction numerator 1 plus straight y plus straight y squared over denominator 1 plus straight y end fraction equals fraction numerator 1 plus 1 plus 1 squared over denominator 1 plus 1 end fraction equals 3 over 2 end style 

Perhatikan bahwa untuk begin mathsize 14px style straight x rightwards arrow 1 end style , sebenarnya didapat pula bahwa begin mathsize 14px style straight y rightwards arrow 1 end style .
Sehingga

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 minus square root of straight y to the power of 6 end root over denominator 1 minus cube root of straight y to the power of 6 end root end fraction end style 

Perhatikan bahwa   begin mathsize 14px style square root of straight y to the power of 6 end root equals open vertical bar straight y cubed close vertical bar end style . Karena begin mathsize 14px style straight y rightwards arrow negative 1 comma space maka space straight y less than 0 comma space sehingga square root of straight y to the power of 6 end root equals negative straight y cubed end style .Maka

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 minus square root of straight y to the power of 6 end root over denominator 1 minus cube root of straight y to the power of 6 end root end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 minus open parentheses negative straight y cubed close parentheses over denominator 1 minus straight y squared end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 plus straight y cubed over denominator 1 minus straight y squared end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator open parentheses 1 plus straight y close parentheses open parentheses 1 minus straight y plus straight y squared close parentheses over denominator open parentheses 1 plus straight y close parentheses open parentheses 1 minus straight y close parentheses end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 minus straight y plus straight y squared over denominator 1 minus straight y end fraction equals fraction numerator 1 minus open parentheses negative 1 close parentheses plus open parentheses negative 1 close parentheses squared over denominator 1 minus open parentheses negative 1 close parentheses end fraction equals fraction numerator 1 plus 1 plus 1 over denominator 1 plus 1 end fraction equals 3 over 2 end style 

Maka

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals 3 over 2 end style 

Jadi, jawabannya adalah D.

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