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Tentukan nilai limit fungsi aljabar dari: x→3lim​(x3+2x2−15x)(x2−3x)​

Pertanyaan

Tentukan nilai limit fungsi aljabar dari:

stack lim space with x rightwards arrow 3 below space fraction numerator open parentheses x squared minus 3 x close parentheses over denominator open parentheses x cubed plus 2 x squared minus 15 x close parentheses end fraction 

Pembahasan Soal:

Limit di atas akan diselesaikan dengan metode pemfaktoran terlebih dahulu karena saat diselesaikan menggunakan metode subtitusi secara langsung ditemukan hasil berbentuk tak tentu 0 over 0.

Dengan memfaktorkan fungsi dalam limit, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell stack lim space with x rightwards arrow 3 below space fraction numerator open parentheses x squared minus 3 x close parentheses over denominator open parentheses x cubed plus 2 x squared minus 15 x close parentheses end fraction end cell equals cell stack lim space with x rightwards arrow 3 below space fraction numerator x open parentheses x minus 3 close parentheses over denominator x open parentheses x squared plus 2 x minus 15 close parentheses end fraction end cell row blank equals cell stack lim space with x rightwards arrow 3 below space fraction numerator open parentheses x minus 3 close parentheses over denominator open parentheses x squared plus 2 x minus 15 close parentheses end fraction end cell row blank equals cell stack lim space with x rightwards arrow 3 below space fraction numerator open parentheses x minus 3 close parentheses over denominator open parentheses x plus 5 close parentheses open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell stack lim space with x rightwards arrow 3 below space fraction numerator 1 over denominator open parentheses x plus 5 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator open parentheses 3 plus 5 close parentheses end fraction end cell row blank equals cell 1 over 8 end cell end table

Jadi, nilai limit fungsi aljabar dari stack lim space with x rightwards arrow 3 below space fraction numerator open parentheses x squared minus 3 x close parentheses over denominator open parentheses x cubed plus 2 x squared minus 15 x close parentheses end fraction adalah 1 over 8.space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Ridha

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Nilai x→7lim​x−72x−14​= ....

Pembahasan Soal:

Substitusi begin mathsize 14px style x equals 7 end style

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

Karena begin mathsize 14px style 0 over 0 end style bentuk tak tentu, maka difaktorkan dahulu

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

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Hitunglah nilai limit fungsi berikut! x→−7lim​x+7x2−49​

Pembahasan Soal:

Dengan menerapkan metode pemfaktoran, limit fungsi di atas dapat diselesaikan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 7 of space fraction numerator x squared minus 49 over denominator x plus 7 end fraction end cell equals cell limit as x rightwards arrow negative 7 of space fraction numerator up diagonal strike open parentheses x plus 7 close parentheses end strike open parentheses x minus 7 close parentheses over denominator up diagonal strike x plus 7 end strike end fraction end cell row blank equals cell limit as x rightwards arrow negative 7 of space open parentheses x minus 7 close parentheses end cell row blank equals cell open parentheses negative 7 close parentheses minus 7 end cell row blank equals cell negative 14 end cell end table end style 

Jadi, nilai dari begin mathsize 14px style limit as x rightwards arrow negative 7 of space fraction numerator x squared minus 49 over denominator x plus 7 end fraction end style adalah begin mathsize 14px style negative 14 end style.

0

Roboguru

Nilaix→1lim​x−12x2−7x+5​=...

Pembahasan Soal:

limit as x rightwards arrow 1 of equals fraction numerator 2 x squared minus 7 x plus 5 over denominator x minus 1 end fraction equals limit as x rightwards arrow 1 of fraction numerator horizontal strike left parenthesis x minus 1 right parenthesis end strike left parenthesis 2 x minus 5 right parenthesis over denominator horizontal strike left parenthesis x minus 1 right parenthesis end strike end fraction equals limit as x rightwards arrow 1 of left parenthesis 2 x minus 5 right parenthesis equals negative 3

0

Roboguru

Jika f(x)=x2−4x2−2x​ maka nilai x→2lim​f(x) adalah ....

Pembahasan Soal:

Diketahui:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight f end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank left parenthesis end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight x end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank right parenthesis 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 cell fraction numerator straight x squared minus 2 straight x over denominator straight x squared minus 4 end fraction end cell end table end style 

Ditanya:

begin mathsize 14px style limit as straight x rightwards arrow 2 of straight f left parenthesis straight x right parenthesis end style

Penyelesaian:

Dengan metode pemfaktoran 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 straight f left parenthesis straight x right parenthesis end cell equals cell limit as straight x rightwards arrow 2 of fraction numerator straight x squared minus 2 straight x over denominator straight x squared minus 4 end fraction end cell row blank equals cell limit as straight x rightwards arrow 2 of fraction numerator straight x left parenthesis straight x minus 2 right parenthesis over denominator left parenthesis straight x plus 2 right parenthesis left parenthesis straight x minus 2 right parenthesis end fraction end cell row blank equals cell limit as straight x rightwards arrow 2 of fraction numerator straight x over denominator left parenthesis straight x plus 2 right parenthesis end fraction end cell row blank equals cell fraction numerator 2 over denominator 2 plus 2 end fraction end cell row blank equals cell 2 over 4 end cell row blank equals cell 1 half end cell end table end style

Jadi, Nilai dari 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 straight f left parenthesis straight x right parenthesis 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 cell 1 half end cell end table end style 

Oleh karena itu, Jawaban yang benar adalah B

0

Roboguru

Nilai x→21​lim​6x2−7x+22x2+5x−3​ adalah ....

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 half of fraction numerator 2 x squared plus 5 x minus 3 over denominator 6 x squared minus 7 x plus 2 end fraction end cell equals cell limit as x rightwards arrow 1 half of fraction numerator open parentheses 2 x minus 1 close parentheses open parentheses x plus 3 close parentheses over denominator open parentheses 2 x minus 1 close parentheses open parentheses 3 x minus 2 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 half of fraction numerator x plus 3 over denominator 3 x minus 2 end fraction end cell row blank equals cell fraction numerator open parentheses begin display style 1 half end style plus 3 close parentheses over denominator 3 open parentheses begin display style 1 half end style close parentheses minus 2 end fraction end cell row blank equals cell fraction numerator begin display style 7 over 2 end style over denominator negative begin display style 1 half end style end fraction end cell row blank equals cell negative 7 end cell end table end style  

Jadi jawaban yang tepat adalah E

0

Roboguru

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