Roboguru

Nilai  adalah...

Nilai begin mathsize 14px style limit as x rightwards arrow negative 4 of space open parentheses fraction numerator 3 over denominator x plus 4 end fraction plus fraction numerator 24 over denominator x squared minus 16 end fraction close parentheses end style adalah...undefined   space  

Jawaban:

Perhatikan penghitungan berikut!

limit as x rightwards arrow negative 4 of space open parentheses fraction numerator 3 over denominator x plus 4 end fraction plus fraction numerator 24 over denominator x squared minus 16 end fraction close parentheses equals limit as x rightwards arrow negative 4 of space open parentheses fraction numerator 3 left parenthesis x minus 4 right parenthesis plus 24 over denominator x squared minus 16 end fraction close parentheses equals limit as x rightwards arrow negative 4 of space open parentheses fraction numerator 3 x minus 12 plus 24 over denominator x squared minus 16 end fraction close parentheses equals limit as x rightwards arrow negative 4 of space open parentheses fraction numerator 3 x plus 12 over denominator x squared minus 16 end fraction close parentheses 

Substitusi:

limit as x rightwards arrow negative 4 of space open parentheses fraction numerator 3 x plus 12 over denominator x squared minus 16 end fraction close parentheses equals fraction numerator 3 left parenthesis negative 4 right parenthesis plus 12 over denominator left parenthesis negative 4 right parenthesis squared minus 16 end fraction equals 0 over 0

Karena hasil substitusi merupakan bentuk tak tentu maka limit akan diselesaikan mengggunakan metode pemfaktoran:

limit as x rightwards arrow negative 4 of space open parentheses fraction numerator 3 x plus 12 over denominator x squared minus 16 end fraction close parentheses equals limit as x rightwards arrow negative 4 of space open parentheses fraction numerator 3 left parenthesis x plus 4 right parenthesis over denominator left parenthesis x minus 4 right parenthesis left parenthesis x plus 4 right parenthesis end fraction close parentheses equals limit as x rightwards arrow negative 4 of space open parentheses fraction numerator 3 over denominator x minus 4 end fraction close parentheses equals fraction numerator 3 over denominator negative 4 minus 4 end fraction equals negative 3 over 8

 

Jadi, limit dari fungsi tersebut adalah negative 3 over 8.undefined 

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