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Tentukan nilai: x → ∞ lim ​ ( 5 x − 1 − 9 x 2 − 18 x + 1 ​ − 4 x 2 + 16 x − 3 ​ )

Tentukan nilai:

 

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N. Puspita

Master Teacher

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 Error converting from MathML to accessible text. 

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Pembahasan

Pembahasan

Dengan menerapkan konsep perkalian sekawan pada limit tak hingga: Kosep limit tak hingga: Jika: Maka: Jadi,

Dengan menerapkan konsep perkalian sekawan pada limit tak hingga:

begin mathsize 12px style limit as straight x rightwards arrow infinity of open parentheses 5 straight x minus 1 minus square root of 9 straight x squared minus 18 straight x plus 1 end root minus square root of 4 straight x squared plus 16 straight x minus 3 end root close parentheses equals limit as x rightwards arrow infinity of open parentheses 3 x plus 2 x minus 1 minus square root of 9 straight x squared minus 18 straight x plus 1 end root minus square root of 4 straight x squared plus 16 straight x minus 3 end root close parentheses equals limit as x rightwards arrow infinity of open parentheses square root of 9 straight x squared end root plus square root of 4 straight x squared end root minus 1 minus square root of 9 straight x squared minus 18 straight x plus 1 end root minus square root of 4 straight x squared plus 16 straight x minus 3 end root close parentheses equals limit as x rightwards arrow infinity of open parentheses square root of 9 straight x squared end root minus square root of 9 straight x squared minus 18 straight x plus 1 end root plus square root of 4 x squared end root minus square root of 4 straight x squared plus 16 straight x minus 3 end root minus 1 close parentheses end style     

Kosep limit tak hingga:

Jika:

begin mathsize 14px style limit as straight x rightwards arrow infinity of left parenthesis square root of ax squared plus bx plus straight c end root minus square root of px squared plus qx plus straight r end root equals open curly brackets table attributes columnalign left end attributes row cell infinity comma space rightwards arrow straight a greater than straight p end cell row cell fraction numerator straight b minus straight q over denominator 2 square root of straight a end fraction rightwards arrow straight a equals straight p end cell row cell negative infinity rightwards arrow straight a less than straight p end cell end table close end style 

Maka:

equals fraction numerator 0 minus left parenthesis negative 18 right parenthesis over denominator 2 square root of 9 end fraction plus fraction numerator 0 minus 16 over denominator 2 square root of 4 end fraction minus 1 equals 18 over 6 plus fraction numerator negative 16 over denominator 4 end fraction minus 1 equals 3 minus 4 minus 1 equals negative 2   

Jadi, Error converting from MathML to accessible text. 

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