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Tentukan nilai dari:   x→2lim​x3−1x2−4​

Pertanyaan

Tentukan nilai dari:

 

begin mathsize 14px style limit as straight x rightwards arrow 2 of fraction numerator straight x squared minus 4 over denominator straight x cubed minus 1 end fraction end style 

Pembahasan Soal:

Mensubstitusikan nilai  begin mathsize 14px style x equals 2 end style maka diperoleh:

  begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 2 of fraction numerator straight x squared minus 4 over denominator straight x cubed minus 1 end fraction end cell equals cell fraction numerator 2 squared minus 4 over denominator 2 cubed minus 1 end fraction end cell row blank equals cell fraction numerator 4 minus 4 over denominator 8 minus 1 end fraction end cell row blank equals cell 0 over 7 end cell row blank equals 0 end table end style 

Jadi, Nilai begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow 2 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator straight x squared minus 4 over denominator straight x cubed minus 1 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 0 end table end style

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. RGFLLIMA

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Tentukan  nilai limit berikut. a. x→4lim​3x−123(x+2)−2(x+5)​

Pembahasan Soal:

Untuk menentukan nilai begin mathsize 14px style limit as x rightwards arrow 4 of space fraction numerator 3 left parenthesis x plus 2 right parenthesis minus 2 left parenthesis x plus 5 right parenthesis over denominator 3 x minus 12 end fraction end style, sederhanakanlah fungsi terlebih dahulu, kemudian substitusikan nilai limitnya.

Ingat!

stack L i m with x rightwards arrow c below space k equals k

Sehingga,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of fraction numerator 3 left parenthesis x plus 2 right parenthesis minus 2 left parenthesis x plus 5 right parenthesis over denominator 3 x minus 12 end fraction end cell equals cell limit as x rightwards arrow 4 of fraction numerator 3 x plus 6 minus 2 x minus 10 over denominator 3 x minus 12 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator x minus 4 over denominator 3 x minus 12 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator x minus 4 over denominator 3 left parenthesis x minus 4 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 4 of 1 third end cell row blank equals cell 1 third end cell end table end style 

Dengan demikian, nilai dari begin mathsize 14px style limit as x rightwards arrow 4 of space fraction numerator 3 left parenthesis x plus 2 right parenthesis minus 2 left parenthesis x plus 5 right parenthesis over denominator 3 x minus 12 end fraction end style adalah 1 third.

0

Roboguru

Nilai x→−4lim​(x+43​+x2−1624​) adalah ...

Pembahasan Soal:

Ada beberapa metode untuk menentukan nilai limit fungsi, jika menggunakan metode substitusi untuk nilai begin mathsize 14px style x equals negative 4 end style maka diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 4 of 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 cell equals cell open parentheses fraction numerator 3 over denominator negative 4 plus 4 end fraction plus fraction numerator 24 over denominator open parentheses negative 4 close parentheses squared minus 16 end fraction close parentheses end cell row blank equals cell open parentheses 3 over 0 plus 24 over 0 close parentheses end cell row blank equals cell 27 over 0 end cell end table end style 

Oleh karena diperoleh hasil begin mathsize 14px style 27 over 0 end style, berdasarkan sifat limit maka gunakan metode pemfaktoran untuk memperoleh nilai dari limit tersebut kemudian substitusikan nilai begin mathsize 14px style x equals negative 4 end style seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 4 of 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 cell equals cell limit as x rightwards arrow negative 4 of open parentheses fraction numerator 3 open parentheses x minus 4 close parentheses over denominator open parentheses x plus 4 close parentheses open parentheses x minus 4 close parentheses end fraction plus fraction numerator 24 over denominator x squared minus 16 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow negative 4 of open parentheses fraction numerator 3 x minus 12 over denominator x squared minus 16 end fraction plus fraction numerator 24 over denominator x squared minus 16 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow negative 4 of open parentheses fraction numerator 3 x plus 12 over denominator x squared minus 16 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow negative 4 of fraction numerator 3 open parentheses x plus 4 close parentheses over denominator open parentheses x plus 4 close parentheses open parentheses x minus 4 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 4 of fraction numerator 3 over denominator x minus 4 end fraction end cell row blank equals cell fraction numerator 3 over denominator negative 4 minus 4 end fraction end cell row blank equals cell fraction numerator 3 over denominator negative 8 end fraction end cell end table end style    

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

x→1lim​(x+3x−2​)sama dengan ....

Pembahasan Soal:

Nilai limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses spacedapat ditentukan dengan mensubstitusi nilai x equals 1 ke dalam fungsi fraction numerator x minus 2 over denominator x plus 3 end fraction sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses end cell equals cell fraction numerator open parentheses 1 close parentheses minus 2 over denominator open parentheses 1 close parentheses plus 3 end fraction end cell row cell limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses end cell equals cell fraction numerator negative 1 over denominator 4 end fraction end cell row cell limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses end cell equals cell negative 1 fourth end cell end table


Jadi nilai limit as x rightwards arrow 1 of open parentheses fraction numerator x minus 2 over denominator x plus 3 end fraction close parentheses spaceadalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell negative 1 fourth. end cell end table

0

Roboguru

x→3lim​x2−2x−3x2−9​=…

Pembahasan Soal:

Dengan menggunakan metode pemfaktoran dan subtitusi limit suatu fungsi, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator x squared minus 9 over denominator x squared minus 2 x minus 3 end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses x plus 3 close parentheses open parentheses x minus 3 close parentheses over denominator open parentheses x minus 3 close parentheses open parentheses x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator x plus 3 over denominator x plus 1 end fraction end cell row blank equals cell fraction numerator left parenthesis 3 right parenthesis plus 3 over denominator left parenthesis 3 right parenthesis plus 1 end fraction end cell row blank equals cell 6 over 4 end cell row blank equals cell 3 over 2 end cell end table 

Dengan demikian, limit as x rightwards arrow 3 of fraction numerator x squared minus 9 over denominator x squared minus 2 x minus 3 end fraction equals 3 over 2.

Jadi, jawaban yang benar adalah E.

0

Roboguru

Diketahui f(x)=x2+3x−4 dan g(x)=2x−7. Tentukan: b.

Pembahasan Soal:

Diketahui Diketahui f open parentheses x close parentheses equals x squared plus 3 x minus 4 dan g open parentheses x close parentheses equals 2 x minus 7. Maka:

limit as x rightwards arrow 3 of space g open parentheses x close parentheses equals limit as x rightwards arrow 3 of space open parentheses 2 x minus 7 close parentheses equals 2 times 3 minus 7 equals 6 minus 7 equals negative 1 

Jadi, nilai limit tersebut adalah negative 1.

0

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