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Nilai dari x → 1 lim ​ ( 2 x − 1 ) 4 x 2 − 4 x 2 ​ adalah...

Nilai dari  adalah...

  1. -e

  2. e

  3. -1

  4. begin mathsize 14px style negative straight e to the power of negative 1 end exponent end style

  5. begin mathsize 14px style straight e to the power of negative 1 end exponent end style

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H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

Perhatikan bahwa Sehingga Misalkan . Sehingga ketika , maka . Maka

Perhatikan bahwa

begin mathsize 14px style limit as straight n rightwards arrow 0 of open parentheses 1 plus straight n close parentheses to the power of 1 over straight n end exponent equals straight e end style 

Sehingga

begin mathsize 14px style limit as straight x rightwards arrow 1 of open parentheses 2 straight x minus 1 close parentheses to the power of fraction numerator 2 over denominator 4 straight x squared minus 4 straight x end fraction end exponent equals limit as straight x rightwards arrow 1 of open parentheses open parentheses 2 straight x minus 1 close parentheses squared close parentheses to the power of fraction numerator 1 over denominator 4 straight x squared minus 4 straight x end fraction end exponent equals limit as straight x rightwards arrow 1 of left parenthesis 4 straight x squared minus 4 straight x plus 1 right parenthesis to the power of fraction numerator 1 over denominator 4 straight x squared minus 4 straight x end fraction end exponent space equals limit as straight x rightwards arrow 1 of left parenthesis 1 minus left parenthesis 4 straight x squared minus 4 straight x right parenthesis right parenthesis to the power of fraction numerator 1 over denominator 4 straight x squared minus 4 straight x end fraction end exponent end style 

Misalkan begin mathsize 14px style straight n equals left parenthesis 4 straight x squared minus 4 straight x right parenthesis end style. Sehingga ketika begin mathsize 14px style straight x rightwards arrow 1 end style, maka undefined.

Maka

begin mathsize 14px style limit as straight x rightwards arrow 1 of open parentheses 1 plus left parenthesis 4 straight x squared minus 4 straight x right parenthesis close parentheses to the power of fraction numerator 1 over denominator 4 straight x squared minus 4 straight x end fraction end exponent equals limit as straight x rightwards arrow 1 of open parentheses 1 minus straight n close parentheses to the power of 1 over straight n end exponent equals straight e end style 

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