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Pertanyaan

limit as x rightwards arrow 2 of open parentheses 2 x plus 5 close parentheses squared equals...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of open parentheses 2 x plus 5 close parentheses squared end cell equals cell open square brackets 2 open parentheses 2 close parentheses plus 5 close square brackets squared end cell row blank equals cell open parentheses 4 plus 5 close parentheses squared end cell row blank equals 81 end table

Jadi, nilai dari Error converting from MathML to accessible text..

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Herlanda

Mahasiswa/Alumni STKIP PGRI Jombang

Terakhir diupdate 02 Mei 2021

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Pertanyaan yang serupa

Nilai = ...

Pembahasan Soal:

limit as x rightwards arrow 25 of fraction numerator 6 square root of x minus 10 over denominator 3 square root of 3 x plus 6 end root minus 2 end fraction

Subtitusikan x equals 25 ke persamaan limit tersebut, sehingga didapat sebagai berikut.

 table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 25 of fraction numerator 6 square root of x minus 10 over denominator 3 square root of 3 x plus 6 end root minus 2 end fraction end cell equals cell fraction numerator 6 square root of 25 minus 10 over denominator 3 square root of 3 times 25 plus 6 end root minus 2 end fraction end cell row blank equals cell fraction numerator 6 times 5 minus 10 over denominator 3 square root of 81 minus 2 end fraction end cell row blank equals cell fraction numerator 30 minus 10 over denominator 3 times 9 minus 2 end fraction end cell row blank equals cell 20 over 25 end cell row blank equals cell 4 over 5 end cell end table 

Jadi, jawaban yang tepat adalah C

0

Roboguru

Pembahasan Soal:

kita coba substitusi begin mathsize 14px style x equals 2 end style, apabila kita mendapatkan hasil yang bukan merupakan bentuk tak tentu. Maka, itulah nilai limitnya.

begin mathsize 14px style space space limit as x rightwards arrow 2 of square root of 2 x squared minus 3 x minus 2 end root equals square root of 2.2 squared minus 3.2 minus 2 end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of 8 minus 6 minus 2 end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0 end style 

Jadi, nilai limitnya adalah 0.

0

Roboguru

....

Pembahasan Soal:

Cara paling dasar dalam menyelesaikan soal limit fungsi aljabar adalah dengan cara substitusi. Apabila hasil dari substitusi tersebut menghasilkan bentuk tak tentu, maka selanjutnya menggunakan cara lain, yaitu dengan faktorisasi dan perkalian sekawan.

Dengan cara substitusi, nilai limit di atas adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of space fraction numerator x minus 2 over denominator square root of x squared plus 1 end root minus square root of 5 end fraction end cell equals cell fraction numerator 2 minus 2 over denominator square root of 2 squared plus 1 end root minus square root of 5 end fraction end cell row blank equals cell 0 over 0 end cell end table

Karena hasil substitusi menghasilkan bentuk tak tentu, sehingga nilai limit akan ditentukan dengan cara perkalian akar sekawan.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 2 of space fraction numerator x minus 2 over denominator square root of x squared plus 1 end root minus square root of 5 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator x minus 2 over denominator square root of x squared plus 1 end root minus square root of 5 end fraction cross times fraction numerator square root of x squared plus 1 end root plus square root of 5 over denominator square root of x squared plus 1 end root plus square root of 5 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses x minus 2 close parentheses open parentheses square root of x squared plus 1 end root plus square root of 5 close parentheses over denominator x squared plus 1 minus 5 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses x minus 2 close parentheses open parentheses square root of x squared plus 1 end root plus square root of 5 close parentheses over denominator x squared minus 4 end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses x minus 2 close parentheses open parentheses square root of x squared plus 1 end root plus square root of 5 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses x plus 2 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses square root of x squared plus 1 end root plus square root of 5 close parentheses over denominator open parentheses x plus 2 close parentheses end fraction end cell row blank equals cell fraction numerator square root of 2 squared plus 1 end root plus square root of 5 over denominator 2 plus 2 end fraction end cell row blank equals cell fraction numerator square root of 5 plus square root of 5 over denominator 4 end fraction end cell row blank equals cell fraction numerator 2 square root of 5 over denominator 4 end fraction end cell row blank equals cell 1 half square root of 5 end cell end table

Dengan demikian, hasil dari table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 2 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x minus 2 over denominator square root of x squared plus 1 end root minus square root of 5 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 5 end cell end table

0

Roboguru

Dengan mensubstitusi nilai-nilai yang didekati oleh , tentukan nilai limit fungsi tersebut. i.

Pembahasan Soal:

Syarat metode substitusi boleh digunakan jika hasil substitusi tidak membentuk nilai bentuk tak tentu seperti begin mathsize 14px style 0 over 0 comma space infinity comma space infinity over infinity comma space infinity minus infinity comma space infinity to the power of infinity comma space 0 to the power of 0 comma space infinity to the power of 0 end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of space fraction numerator x to the power of 4 minus 1 over denominator x squared minus 1 end fraction end cell equals cell fraction numerator 2 to the power of 4 minus 1 over denominator 2 squared minus 1 end fraction end cell row blank equals cell 15 over 3 end cell row blank equals 5 end table end style

Jadi, nilai begin mathsize 14px style limit as x rightwards arrow 2 of space fraction numerator x to the power of 4 minus 1 over denominator x squared minus 1 end fraction equals 5 end style.

0

Roboguru

Tentukan nilai limit fungsi berikut.

Pembahasan Soal:

Diketahui:

Limit fungsi akar kuadrat.

Substitusi begin mathsize 14px style x end style dengan begin mathsize 14px style 9 end style pada fungsi limit akar kuadrat seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 9 of space square root of 4 x minus 11 end root end cell equals cell square root of 4 open parentheses 9 close parentheses minus 11 end root end cell row blank equals cell square root of 36 minus 11 end root end cell row blank equals cell square root of 25 end cell row blank equals 5 end table end style

Maka, nilai dari begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 9 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 4 x minus 11 end root 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 5 end table end style.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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