Tentukan nilai limit fungsi berikut!

Pertanyaan

$begin mathsize 14px style limit as x rightwards arrow 2 of space fraction numerator square root of 5 x minus 6 end root minus x over denominator x squared plus x minus 6 end fraction end style$

Pembahasan Soal:

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

$x \to 2 lim \frac{5 x - 6 - x}{x ^{2} + x - 6} = \frac{5 ( 2 ) - 6 - x}{2 ^{2} + 2 - 6} = \frac{0}{0}$

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 $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell size 14px 1 over size 14px 20 end cell end table$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Firmansyah

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

Nilai dari

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.

Roboguru

Pembahasan Soal:

Dengan mengalikan akar sekawan, diperoleh perhitungan sebagai berikut:

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

Jadi, diperoleh nilai $begin mathsize 14px style limit as x rightwards arrow 1 of fraction numerator 1 minus square root of x over denominator 1 minus x squared end fraction equals 1 fourth end style$ .

Roboguru

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

Roboguru

Tentukan nilai dari

Pembahasan Soal:

Subtitusi nilai x:

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

Roboguru

Nilai dari adalah

Pembahasan Soal:

Jika kita menghitung nilai $begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator open parentheses x squared minus 1 close parentheses open parentheses x squared minus 4 x close parentheses over denominator x squared minus x end fraction end style$ dengan cara substitusi langsung, maka kita akan memperoleh bentuk tak tentu $begin mathsize 14px style 0 over 0 end style$ . Oleh karena itu, limit fungsi tersebut hendaknya diselesaikan dengan cara pemfaktoran, yaitu

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

Dengan demikian, $begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as p rightwards arrow negative 2 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator begin display style 4 minus p squared end style over denominator 2 plus p end fraction end cell end table equals 4 end style$ $begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator left parenthesis x squared minus 1 right parenthesis left parenthesis x squared minus 4 x right parenthesis over denominator x squared minus x end fraction equals negative 4 end style$ .

Jadi, jawaban yang tepat adalah A.

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