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x→3lim​x−4x−3​5x−15​=…

Pertanyaan

begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 5 x minus 15 over denominator x minus square root of 4 x minus 3 end root end fraction equals horizontal ellipsis end style

  1. begin mathsize 14px style horizontal ellipsis end style

  2. begin mathsize 14px style horizontal ellipsis end style

Pembahasan Soal:

Jika kita melakukan substitusi langsung pada permasalahan limit di atas, kita akan memperoleh bentuk tak tentu begin mathsize 14px style 0 over 0 end style. Oleh karena itu, kita akan menyelesaikan masalah limit tersebut dengan cara mengalikan akar sekawan penyebutnya.

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator 5 x minus 15 over denominator x minus square root of 4 x minus 3 end root end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator 5 x minus 15 over denominator x minus square root of 4 x minus 3 end root end fraction times fraction numerator x plus square root of 4 x minus 3 end root over denominator x plus square root of 4 x minus 3 end root end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 5 x minus 15 close parentheses open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator x squared minus open parentheses square root of 4 x minus 3 end root close parentheses squared end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 5 x minus 15 close parentheses open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator x squared minus open parentheses 4 x minus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses 5 x minus 15 close parentheses open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator x squared minus 4 x plus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator 5 up diagonal strike open parentheses x minus 3 close parentheses end strike open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator up diagonal strike open parentheses x minus 3 close parentheses end strike open parentheses x minus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator 5 open parentheses x plus square root of 4 x minus 3 end root close parentheses over denominator open parentheses x minus 1 close parentheses end fraction end cell row blank equals cell fraction numerator 5 open parentheses 3 plus square root of 4 times 3 minus 3 end root close parentheses over denominator open parentheses 3 minus 1 close parentheses end fraction end cell row blank equals cell fraction numerator 5 open parentheses 3 plus square root of 9 close parentheses over denominator open parentheses 2 close parentheses end fraction end cell row blank equals cell fraction numerator 5 open parentheses 3 plus 3 close parentheses over denominator 2 end fraction end cell row blank equals 15 end table end style

Dengan demikian, begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator 5 x minus 15 over denominator x minus square root of 4 x minus 3 end root end fraction equals 15 end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Tentukan nilai limit fungsi berikut! x→2lim​x2+x−65x−6​−x​

Pembahasan Soal:

Langkah pertama untuk mencari nilai limit xa dapat dilakukan dengan mensubstitusikan nilai x=a ke bentuk aljabarnya. Perhatikan perhitungan berikut

x2limx2+x65x6x=22+265(2)6x=00

Karena menghasilkan bentuk tak tentu maka dapat dicari dengan cara lain.

Dengan menerapkan konsep penyelesaian limit fungsi dengan metode mengalikan akar sekawan dan pemfaktoran, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 2 of space fraction numerator square root of 5 x minus 6 end root minus x over denominator x squared plus x minus 6 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator square root of 5 x minus 6 end root minus x over denominator x squared plus x minus 6 end fraction cross times fraction numerator square root of 5 x minus 6 end root plus x over denominator square root of 5 x minus 6 end root plus x end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses 5 x minus 6 close parentheses minus x squared over denominator open parentheses x squared plus x minus 6 close parentheses open parentheses square root of 5 x minus 6 end root plus x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator negative x squared plus 5 x minus 6 over denominator open parentheses x squared plus x minus 6 close parentheses open parentheses square root of 5 x minus 6 end root plus x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator negative open parentheses x squared minus 5 x plus 6 close parentheses over denominator open parentheses x squared plus x minus 6 close parentheses open parentheses square root of 5 x minus 6 end root plus x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator negative up diagonal strike open parentheses x minus 2 close parentheses end strike open parentheses x minus 3 close parentheses over denominator up diagonal strike open parentheses x minus 2 close parentheses end strike open parentheses x plus 3 close parentheses open parentheses square root of 5 x minus 6 end root plus x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator negative open parentheses x minus 3 close parentheses over denominator open parentheses x plus 3 close parentheses open parentheses square root of 5 x minus 6 end root plus x close parentheses end fraction end cell row blank equals cell fraction numerator negative open parentheses 2 minus 3 close parentheses over denominator open parentheses 2 plus 3 close parentheses open parentheses square root of 5 open parentheses 2 close parentheses minus 6 end root plus 2 close parentheses end fraction end cell row blank equals cell fraction numerator negative open parentheses negative 1 close parentheses over denominator 5 open parentheses square root of 4 plus 2 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator 5 open parentheses 4 close parentheses end fraction end cell row blank equals cell 1 over 20 end cell end table end style 

Jadi, nilai begin mathsize 14px style limit as x rightwards arrow 2 of space fraction numerator square root of 5 x minus 6 end root minus x over denominator x squared plus x minus 6 end fraction end style adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell size 14px 1 over size 14px 20 end cell end table.

0

Roboguru

Nilai dari x→1lim​x2+3​−2x−1​=…

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator x minus 1 over denominator square root of x squared plus 3 end root minus 2 end fraction end cell equals cell limit as x rightwards arrow 1 of fraction numerator x minus 1 over denominator square root of x squared plus 3 end root minus 2 end fraction cross times fraction numerator square root of x squared plus 3 end root plus 2 over denominator square root of x squared plus 3 end root plus 2 end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator open parentheses x minus 1 close parentheses open parentheses square root of x squared plus 3 end root plus 2 close parentheses over denominator open parentheses x squared plus 3 close parentheses minus 4 end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator left parenthesis x minus 1 right parenthesis open parentheses square root of x squared plus 3 end root plus 2 close parentheses over denominator x squared minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator left parenthesis x minus 1 right parenthesis open parentheses square root of x squared plus 3 end root plus 2 close parentheses over denominator open parentheses x minus 1 close parentheses open parentheses x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator up diagonal strike left parenthesis x minus 1 right parenthesis end strike open parentheses square root of x squared plus 3 end root plus 2 close parentheses over denominator up diagonal strike open parentheses x minus 1 close parentheses end strike open parentheses x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator square root of x squared plus 3 end root plus 2 over denominator x plus 1 end fraction end cell row blank equals cell fraction numerator square root of 1 squared plus 3 end root plus 2 over denominator 1 plus 1 end fraction end cell row blank equals cell fraction numerator 2 plus 2 over denominator 2 end fraction end cell row blank equals 2 row blank blank blank end table end style  

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Tentukan nilai dari x→0lim​x2+x​−2−x​​

Pembahasan Soal:

Subtitusi nilai x:

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

Karena hasil merupakan bentuk tak tentu, maka difaktorkan sebagai berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator square root of 2 plus x end root minus square root of 2 minus x end root over denominator x end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator square root of 2 plus x end root minus square root of 2 minus x end root over denominator x end fraction times fraction numerator square root of 2 plus x end root plus square root of 2 minus x end root over denominator square root of 2 plus x end root plus square root of 2 minus x end root end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator left parenthesis 2 plus x right parenthesis minus left parenthesis 2 minus x right parenthesis over denominator x left parenthesis square root of 2 plus x end root plus square root of 2 minus x end root right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 2 x over denominator x left parenthesis square root of 2 plus x end root plus square root of 2 minus x end root right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 2 over denominator square root of 2 plus x end root plus square root of 2 minus x end root end fraction end cell row blank equals cell fraction numerator 2 over denominator square root of 2 plus 0 end root plus square root of 2 minus 0 end root end fraction end cell row blank equals cell fraction numerator 2 over denominator 2 square root of 2 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 2 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 2 end fraction times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals cell 1 half square root of 2 end cell end table end style 

Jadi, begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 0 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator square root of 2 plus x end root minus square root of 2 minus x end root over denominator x 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 cell 1 half end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank end table end style.

0

Roboguru

Nilaidarix→8lim​3x−2​x−8​=....

Pembahasan Soal:

limit as x rightwards arrow 8 of fraction numerator x minus 8 over denominator cube root of x minus 2 end root end fraction equals fraction numerator 8 minus 0 over denominator cube root of 8 minus 2 end root end fraction equals 0 over 0  J i k a space h a s i l space l i m i t space 0 over 0 comma space m a k a space m e n u r u t space t e o r i space L apostrophe h o s p i t a l space b e n t u k space l i m i t space d a p a t space d i s e l e s a i k a n space d e n g a n space m e n d i f e r e n s i a l k a n space f u n g sin y a.  limit as x rightwards arrow 8 of fraction numerator x minus 8 over denominator cube root of x minus 2 end root end fraction equals limit as x rightwards arrow 8 of fraction numerator 1 over denominator begin display style 3 over 4 end style left parenthesis 8 right parenthesis cube root of x end fraction equals fraction numerator 1 over denominator begin display style 3 over 4 left parenthesis 8 right parenthesis cube root of x end style end fraction equals 12

0

Roboguru

Tentukan nilai dari:  x→16lim​x−16x​−4​

Pembahasan Soal:

Cara perkalian akar sekawan dipakai jika hasil uji substitusi menghasilkan bentuk tak tentu, dan khusus untuk soal limit yang fungsinya berbentuk akar.

Pertama, kita uji terlebih dahulu dengan menggunakan cara substitusi.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 16 of fraction numerator square root of x minus 4 over denominator x minus 16 end fraction end cell equals cell fraction numerator square root of 16 minus 4 over denominator 16 minus 16 end fraction end cell row blank equals cell 0 over 0 end cell end table 

Karena menghasilkan bentuk tak tentu, maka kita gunakan cara perkalian akar sekawan seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 16 of fraction numerator square root of x minus 4 over denominator x minus 16 end fraction end cell equals cell limit as x rightwards arrow 16 of open parentheses fraction numerator square root of x minus 4 over denominator x minus 16 end fraction close parentheses times open parentheses fraction numerator square root of x plus 4 over denominator square root of x plus 4 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 16 of fraction numerator open parentheses x minus 16 close parentheses over denominator open parentheses x minus 16 close parentheses open parentheses square root of x plus 4 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 16 of fraction numerator 1 over denominator square root of x plus 4 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 16 plus 4 end fraction end cell row blank equals cell fraction numerator 1 over denominator 4 plus 4 end fraction end cell row blank equals cell 1 over 8 end cell end table end style  

Jadi begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 16 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator square root of x minus 4 over denominator x minus 16 end fraction end cell end table equals 1 over 8 end style

0

Roboguru

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