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Dengan cara substitusi  d.

Pertanyaan

Dengan cara substitusi 

d. limit as x rightwards arrow 5 of space 7 minus 3 x equals..

Pembahasan Soal:

Dengan cara substitusi 

d. limit as x rightwards arrow 5 of space 7 minus 3 x equals..

Penyelesaian 

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 5 of space 7 minus 3 straight x end cell equals cell 7 minus 3 open parentheses 5 close parentheses end cell row blank equals cell 7 minus 15 end cell row blank equals cell negative 8 end cell end table

jadi, nilai dari limit as x rightwards arrow 5 of space 7 minus 3 x equals negative 8

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Rahmawati

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 06 Juni 2021

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Pertanyaan yang serupa

Tunjukkan dengan pendekatan nilai  !

Pembahasan Soal:

Nilai begin mathsize 14px style 6 x cubed end stylebegin mathsize 14px style 2 x end style, dan begin mathsize 14px style 3 x squared end style pada saat undefined mendekati undefined.

Jadi,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis limit as x rightwards arrow 2 of 2 x right parenthesis left parenthesis limit as x rightwards arrow 2 of 3 x squared right parenthesis end cell equals cell left parenthesis 4 right parenthesis left parenthesis 12 right parenthesis end cell row blank equals 48 row blank equals cell limit as x not stretchy rightwards arrow 2 of 6 x cubed. end cell end table end style

0

Roboguru

Jika  dan , maka

Pembahasan Soal:

begin mathsize 12px style limit as x rightwards arrow a of open parentheses f open parentheses x close parentheses plus fraction numerator 1 over denominator g open parentheses x close parentheses end fraction close parentheses equals 4  limit as x rightwards arrow a of f open parentheses x close parentheses plus limit as x rightwards arrow a of fraction numerator 1 over denominator g open parentheses x close parentheses end fraction equals negative 3  Misalkan colon  limit as straight x rightwards arrow straight a of straight f open parentheses straight x close parentheses equals straight p space dan space limit as straight x rightwards arrow straight a of fraction numerator 1 over denominator straight g open parentheses straight x close parentheses end fraction equals straight q  Maka space diperoleh colon  straight p plus straight q equals 4    limit as straight x rightwards arrow straight a of open parentheses straight f open parentheses straight x close parentheses minus fraction numerator 1 over denominator straight g open parentheses straight x close parentheses end fraction close parentheses equals 4  limit as straight x rightwards arrow straight a of straight f open parentheses straight x close parentheses minus limit as straight x rightwards arrow straight a of fraction numerator 1 over denominator straight g open parentheses straight x close parentheses end fraction equals negative 3  Misalkan colon  limit as straight x rightwards arrow straight a of straight f open parentheses straight x close parentheses equals straight p space dan space limit as straight x rightwards arrow straight a of fraction numerator 1 over denominator straight g open parentheses straight x close parentheses end fraction equals straight q  Maka space diperoleh colon  straight p minus straight q equals negative 3    Eliminasi colon  straight p plus straight q equals 4  bottom enclose straight p minus straight q equals negative 3 end enclose space minus  2 straight q equals 7  straight q equals 7 over 2  limit as straight x rightwards arrow straight a of fraction numerator 1 over denominator straight g open parentheses straight x close parentheses end fraction equals 7 over 2  limit as straight x rightwards arrow straight a of straight g open parentheses straight x close parentheses equals 2 over 7    straight p plus straight q equals 4  straight p plus 7 over 2 equals 4  straight p equals 1 half  limit as straight x rightwards arrow straight a of straight f open parentheses straight x close parentheses equals 1 half    Sehingga colon  limit as straight x rightwards arrow straight a of straight f open parentheses straight x close parentheses straight g open parentheses straight x close parentheses equals limit as straight x rightwards arrow straight a of straight f open parentheses straight x close parentheses times limit as straight x rightwards arrow straight a of straight g open parentheses straight x close parentheses  limit as straight x rightwards arrow straight a of straight f open parentheses straight x close parentheses straight g open parentheses straight x close parentheses equals 1 half times 2 over 7 equals 2 over 14 end style

1

Roboguru

Dengan menggunakan sifat-sifat limit fungsi aljabar, tentukan nilai .

Pembahasan Soal:

Dalam limit tersebut kita gunakan sifat limit begin mathsize 14px style limit as x rightwards arrow c of open square brackets f left parenthesis x right parenthesis. g left parenthesis x right parenthesis close square brackets equals limit as x rightwards arrow c of f left parenthesis x right parenthesis. limit as x rightwards arrow c of g left parenthesis x right parenthesis end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of open parentheses 4 x squared plus 5 x plus 1 close parentheses end cell equals cell limit as x rightwards arrow 4 of open parentheses 4 x plus 1 close parentheses left parenthesis x plus 1 right parenthesis end cell row blank equals cell limit as x rightwards arrow 4 of open parentheses 4 x plus 1 close parentheses times limit as x rightwards arrow 4 of left parenthesis x plus 1 right parenthesis end cell row blank equals cell left parenthesis 4 left parenthesis 4 right parenthesis plus 1 right parenthesis times left parenthesis 4 plus 1 right parenthesis end cell row blank equals cell left parenthesis 16 plus 1 right parenthesis times left parenthesis 5 right parenthesis end cell row blank equals cell 17 times 5 end cell row blank equals 85 end table end style

Jadi, nilai limit begin mathsize 14px style limit as x rightwards arrow 4 of open parentheses 4 x squared plus 5 x plus 1 close parentheses end style adalah begin mathsize 14px style 85 end style.

0

Roboguru

Diketahui , , dan . Tentukan: .

Pembahasan Soal:

Ingat sifat limit bahwa begin mathsize 14px style limit as x rightwards arrow a of fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals fraction numerator limit as x rightwards arrow a of f left parenthesis x right parenthesis over denominator limit as x rightwards arrow a of g left parenthesis x right parenthesis end fraction end stylebegin mathsize 14px style limit as x rightwards arrow a of left parenthesis f left parenthesis x right parenthesis times g left parenthesis x right parenthesis right parenthesis equals limit as x rightwards arrow a of f left parenthesis x right parenthesis times limit as x rightwards arrow a of g left parenthesis x right parenthesis end style dan begin mathsize 14px style limit as x rightwards arrow a of left parenthesis f left parenthesis x right parenthesis plus-or-minus g left parenthesis x right parenthesis right parenthesis equals limit as x rightwards arrow a of f left parenthesis x right parenthesis plus-or-minus limit as x rightwards arrow a of g left parenthesis x right parenthesis end style, maka

begin mathsize 14px style limit as x rightwards arrow a of fraction numerator h left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction left parenthesis f left parenthesis x right parenthesis times g left parenthesis x right parenthesis plus 2 h left parenthesis x right parenthesis right parenthesis equals limit as x rightwards arrow a of fraction numerator h left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction times limit as x rightwards arrow a of left parenthesis f left parenthesis x right parenthesis times g left parenthesis x right parenthesis plus 2 h left parenthesis x right parenthesis right parenthesis equals fraction numerator limit as x rightwards arrow a of h left parenthesis x right parenthesis over denominator limit as x rightwards arrow a of f left parenthesis x right parenthesis end fraction times open parentheses limit as x rightwards arrow a of left parenthesis f left parenthesis x right parenthesis times g left parenthesis x right parenthesis right parenthesis plus limit as x rightwards arrow a of 2 h left parenthesis x right parenthesis close parentheses equals fraction numerator limit as x rightwards arrow a of h left parenthesis x right parenthesis over denominator limit as x rightwards arrow a of f left parenthesis x right parenthesis end fraction times open parentheses limit as x rightwards arrow a of left parenthesis f left parenthesis x right parenthesis times limit as x rightwards arrow a of g left parenthesis x right parenthesis right parenthesis plus stack 2 times lim with x rightwards arrow a below h left parenthesis x right parenthesis close parentheses equals 4 over 8 open parentheses 8 times left parenthesis negative 3 right parenthesis plus 2 times 4 close parentheses equals 1 half left parenthesis negative 24 plus 8 right parenthesis equals 1 half left parenthesis negative 16 right parenthesis equals negative 8 end style

Jadi hasil dari begin mathsize 14px style limit as x rightwards arrow a of fraction numerator h left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction left parenthesis f left parenthesis x right parenthesis times g left parenthesis x right parenthesis plus 2 h left parenthesis x right parenthesis right parenthesis end style adalah begin mathsize 14px style negative 8 end style.

0

Roboguru

Diketahui dan . Nilai dari adalah

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 3 of open square brackets f open parentheses x close parentheses times g open parentheses x close parentheses minus x close square brackets end cell row blank equals cell limit as x rightwards arrow 3 of f open parentheses x close parentheses times limit as x rightwards arrow 3 of g open parentheses x close parentheses minus limit as x rightwards arrow 3 of x end cell row blank equals cell negative 2 times 5 1 half minus 3 end cell row blank equals cell negative 11 minus 3 end cell row blank equals cell negative 14 end cell end table end style

Jadi, jawaban yang tepat adalah E.

0

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