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Tentukan nilai: x → ∞ lim ​ ( 2 x − 2 − 4 x 2 − 4 x + 3 ​ )

Tentukan nilai:

 

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N. Puspita

Master Teacher

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nilai dari .

nilai dari limit as x rightwards arrow infinity of left parenthesis 2 straight x minus 2 minus square root of 4 straight x squared minus 4 straight x plus 3 end root right parenthesis equals negative 1 .

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Ingat kembali : jika dan hanya jika jika dan hanya jika jika dan hanya jika sehingga karna nilai maka Jadi, nilai dari .

Ingat kembali :

begin mathsize 14px style limit as x rightwards arrow infinity of open parentheses square root of a x squared plus b x plus c end root minus square root of p x squared plus q x plus r end root close parentheses equals L end style 

  • begin mathsize 14px style L equals negative infinity end style jika dan hanya jika begin mathsize 14px style a less than p end style
  • L equals fraction numerator b minus q over denominator 2 square root of a end fraction jika dan hanya jika begin mathsize 14px style a equals p end style
  • begin mathsize 14px style L equals infinity end style jika dan hanya jika begin mathsize 14px style a greater than p end style

 

sehingga 

begin mathsize 12px style limit as x rightwards arrow infinity of left parenthesis 2 straight x minus 2 minus square root of 4 straight x squared minus 4 straight x plus 3 end root right parenthesis equals limit as x rightwards arrow infinity of open parentheses square root of left parenthesis 2 x minus 2 right parenthesis squared end root minus square root of 4 straight x squared minus 4 straight x plus 3 end root close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals limit as straight x rightwards arrow infinity of open parentheses square root of 4 straight x squared minus 8 straight x plus 4 end root minus square root of 4 straight x squared minus 4 straight x plus 3 end root close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals L end style

karna nilai a equals p equals 4 maka 

L equals fraction numerator b minus q over denominator 2 square root of a end fraction space space space equals fraction numerator negative 8 minus left parenthesis negative 4 right parenthesis over denominator 2 square root of 4 end fraction space space space equals fraction numerator negative 8 plus 4 over denominator 2 times 2 end fraction equals fraction numerator negative 4 over denominator 4 end fraction equals negative 1

 

Jadi, nilai dari limit as x rightwards arrow infinity of left parenthesis 2 straight x minus 2 minus square root of 4 straight x squared minus 4 straight x plus 3 end root right parenthesis equals negative 1 .

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