Nilai dari  adalah ....

Pertanyaan

Nilai dari begin mathsize 14px style lim subscript straight x rightwards arrow 3 end subscript open parentheses fraction numerator 2 over denominator straight x minus 3 end fraction minus fraction numerator 12 over denominator straight x squared minus 9 end fraction close parentheses end style adalah ....

  1. begin mathsize 14px style minus end stylebegin mathsize 14px style 1 over 6 end style 

  2. begin mathsize 14px style minus end stylebegin mathsize 14px style 1 third end style 

  3. begin mathsize 14px style 1 third end style  

  4. begin mathsize 14px style 1 over 6 end style 

  5. begin mathsize 14px style straight infinity end style 

R. RGFLSATU

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Perhatikan bahwa jika dilakukan substitusi = 3, maka didapatkan hasil begin mathsize 14px style 2 over 0 minus 12 over 0 end style. Bentuk ini merupakan salah satu bentuk tak tentu yang tidak dapat dicari secara langsung penyelesaiannya. Sehingga perlu dicari cara lain untuk mencari nilai limitnya.

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow 3 of open parentheses fraction numerator 2 over denominator straight x minus 3 end fraction minus fraction numerator 12 over denominator straight x squared minus 9 end fraction close parentheses end cell row blank equals cell limit as straight x rightwards arrow 3 of open parentheses fraction numerator 2 over denominator straight x minus 3 end fraction times fraction numerator straight x plus 3 over denominator straight x plus 3 end fraction minus fraction numerator 12 over denominator straight x squared minus 9 end fraction close parentheses end cell row blank equals cell limit as straight x rightwards arrow 3 of open parentheses fraction numerator 2 open parentheses straight x plus 3 close parentheses over denominator straight x squared minus 9 end fraction minus fraction numerator 12 over denominator straight x squared minus 9 end fraction close parentheses end cell row blank equals cell limit as straight x rightwards arrow 3 of open parentheses fraction numerator 2 straight x plus 6 over denominator straight x squared minus 9 end fraction minus fraction numerator 12 over denominator straight x squared minus 9 end fraction close parentheses end cell row blank equals cell limit as straight x rightwards arrow 3 of fraction numerator 2 straight x plus 6 minus 12 over denominator straight x squared minus 9 end fraction end cell row blank equals cell limit as straight x rightwards arrow 3 of fraction numerator 2 straight x minus 6 over denominator straight x squared minus 9 end fraction end cell row blank equals cell limit as straight x rightwards arrow 3 of fraction numerator 2 open parentheses straight x minus 3 close parentheses over denominator open parentheses straight x plus 3 close parentheses open parentheses straight x minus 3 close parentheses end fraction end cell row blank equals cell limit as straight x rightwards arrow 3 of fraction numerator 2 over denominator straight x plus 3 end fraction end cell row blank equals cell fraction numerator 2 over denominator 3 plus 3 end fraction end cell row blank equals cell 2 over 6 end cell row blank equals cell 1 third end cell end table end style 

Jadi, jawaban yang tepat adalah C.

97

5.0 (6 rating)

Pertanyaan serupa

limx→4​3−x+5​4−x+5​​−x+5​−2​​= ....

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Jawaban terverifikasi

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