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Pertanyaan

begin mathsize 14px style limit as x rightwards arrow 4 of space 2 space plus limit as x rightwards arrow 3 of space 8 space equals space... end style 

Pembahasan Video:

Pembahasan Soal:

Dengan menggunakan sifat limit begin mathsize 14px style limit as x rightwards arrow c of k equals k end style, maka pertanyaan di atas dapat diselesaikan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell limit as x rightwards arrow 4 of 2 plus limit as x rightwards arrow 3 of 8 end cell equals cell 2 plus 8 end cell row blank equals 10 end table end style

Dengan demikian, begin mathsize 14px style limit as x rightwards arrow 4 of 2 plus limit as x rightwards arrow 3 of 8 equals 10 end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Sibuea

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 13 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai  adalah...

Pembahasan Soal:

Dengan menggunakan konsep dasar limit diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 5 of left parenthesis 3 x squared minus 15 x plus 10 right parenthesis end cell equals cell 3 open parentheses 5 close parentheses squared minus 15 open parentheses 5 close parentheses plus 10 end cell row blank equals cell 3 open parentheses 25 close parentheses minus 75 plus 10 end cell row blank equals cell 75 minus 75 plus 10 end cell row blank equals 10 end table end style

Dengan demikian nilai dari begin mathsize 14px style limit as x rightwards arrow 5 of left parenthesis 3 x squared minus 15 x plus 10 right parenthesis end style adalah begin mathsize 14px style 10 end style.

1

Roboguru

Tentukan nilai limit berikut: 1.

Pembahasan Soal:

Gunakan konsep limit bentuk aljabar dan limit pada fungsi pecahan.

Perhatikan perhitungan berikut.

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

Jadi, diperoleh begin mathsize 14px style limit as x rightwards arrow 1 of space fraction numerator x squared plus 1 over denominator x minus 2 end fraction equals negative 2 end style.

0

Roboguru

Pembahasan Soal:

Kita gunakan metode substitusi. 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 negative 5 of open parentheses 2 x to the power of 4 plus 3 x cubed minus 25 close parentheses end cell equals cell 2 open parentheses negative 5 close parentheses to the power of 4 plus 3 open parentheses negative 5 close parentheses cubed minus 25 end cell row blank equals cell 1.250 plus open parentheses negative 375 close parentheses minus 25 end cell row blank equals 850 end table end style

Jadi, nilai begin mathsize 14px style limit as x rightwards arrow negative 5 of space open parentheses 2 x to the power of 4 plus 3 x cubed minus 25 close parentheses equals 850 end style.

1

Roboguru

Dengan teorema limit, hitunglah :

Pembahasan Soal:

Dengan menggunakan konsep :
 

begin mathsize 14px style limit as straight x rightwards arrow straight c of open parentheses straight f open parentheses straight x close parentheses cross times straight g open parentheses straight x close parentheses close parentheses equals limit as straight x rightwards arrow straight c of open parentheses straight f open parentheses straight x close parentheses close parentheses cross times limit as straight x rightwards arrow straight c of open parentheses straight g open parentheses straight x close parentheses close parentheses end style 
 

maka pada begin mathsize 14px style limit as straight x rightwards arrow negative 2 of open parentheses straight x plus 4 close parentheses open parentheses straight x squared plus 3 straight x minus 5 close parentheses end style diubah bentuk menjadi
 

begin mathsize 14px style limit as straight x rightwards arrow negative 2 of open parentheses straight x plus 4 close parentheses cross times limit as straight x rightwards arrow negative 2 of open parentheses straight x squared plus 3 straight x minus 5 close parentheses end style 

Kemudian ingat bahwa

 

begin mathsize 14px style limit as straight x rightwards arrow straight c of open parentheses straight f open parentheses straight x close parentheses plus-or-minus straight g open parentheses straight x close parentheses close parentheses equals limit as straight x rightwards arrow straight c of straight f open parentheses straight x close parentheses plus-or-minus limit as straight x rightwards arrow straight c of straight g open parentheses straight x close parentheses end style 
 

Maka :

 

begin mathsize 14px style limit as straight x rightwards arrow negative 2 of open parentheses straight x plus 4 close parentheses cross times limit as straight x rightwards arrow negative 2 of open parentheses straight x squared plus 3 straight x minus 5 close parentheses equals open parentheses limit as straight x rightwards arrow negative 2 of open parentheses straight x close parentheses plus limit as straight x rightwards arrow negative 2 of open parentheses 4 close parentheses close parentheses cross times open parentheses limit as straight x rightwards arrow negative 2 of open parentheses straight x squared close parentheses plus limit as straight x rightwards arrow negative 2 of open parentheses 3 straight x close parentheses plus limit as straight x rightwards arrow negative 2 of open parentheses 5 close parentheses close parentheses equals open parentheses negative 2 plus 4 close parentheses cross times open parentheses open parentheses negative 2 close parentheses squared plus 3 open parentheses negative 2 close parentheses minus 5 close parentheses equals 2 cross times open parentheses 4 minus 6 minus 5 close parentheses equals 2 cross times open parentheses negative 7 close parentheses equals negative 14 end style 
 

Jadi, begin mathsize 14px style limit as straight x rightwards arrow negative 2 of open parentheses straight x plus 4 close parentheses open parentheses straight x squared plus 3 straight x minus 5 close parentheses equals negative 14 end style 

0

Roboguru

Pembahasan Soal:

Ingat sifat limit : begin mathsize 14px style limit as x rightwards arrow c of k equals k end style

Sehingga kita dapat begin mathsize 14px style limit as x rightwards arrow 3 of 4 equals 4 end style.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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