Roboguru

x→0lim​sin4xtan3x7x2​=...

Pertanyaan

limit as x rightwards arrow 0 of space fraction numerator 7 x squared over denominator sin space 4 x space tan space 3 x end fraction equals...

  1. 0

  2. 7 over 12

  3. 1

  4. 12 over 7

  5. infinity

Pembahasan Soal:

Ingat kembali undefined maka:

limit as x rightwards arrow 0 of space fraction numerator 7 x squared over denominator sin space 4 x space tan space 3 x end fraction  equals limit as x rightwards arrow 0 of space open parentheses 7 close parentheses open parentheses fraction numerator x over denominator sin space 4 x end fraction close parentheses open parentheses fraction numerator x over denominator tan space 3 x end fraction close parentheses  equals open parentheses 7 close parentheses open parentheses 1 fourth close parentheses open parentheses 1 third close parentheses  equals 7 over 12

Jawaban B

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Roy

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 04 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

x→4π​lim​1−tanxsinxcosx​=...

Pembahasan Soal:

Dengan menerapkan penyelesaian limit dengan metode substitusi, diperoleh hasil perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow pi over 4 of space fraction numerator sin space x space cos space x over denominator 1 minus tan space x end fraction end cell equals cell fraction numerator sin space open parentheses begin display style pi over 4 end style close parentheses space cos space open parentheses begin display style pi over 4 end style close parentheses over denominator 1 minus tan space open parentheses begin display style pi over 4 end style close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses begin display style 1 half end style square root of 2 close parentheses open parentheses begin display style 1 half end style square root of 2 close parentheses over denominator 1 minus 1 end fraction end cell row blank equals cell fraction numerator begin display style 1 half end style over denominator 0 end fraction end cell row blank equals infinity end table 

Jadi, nilai limit as x rightwards arrow pi over 4 of space fraction numerator sin space x space cos space x over denominator 1 minus tan space x end fraction adalah infinity.

1

Roboguru

x→0lim​1−sec22x2xsin3x​=⋯

Pembahasan Soal:

size 14px lim with size 14px x size 14px rightwards arrow size 14px 0 below size 14px space fraction numerator size 14px 2 size 14px x size 14px space size 14px sin size 14px space size 14px 3 size 14px x over denominator size 14px 1 size 14px minus size 14px s size 14px e size 14px c to the power of size 14px 2 size 14px space size 14px 2 size 14px x end fraction size 14px equals size 14px lim with size 14px x size 14px rightwards arrow size 14px 0 below size 14px space fraction numerator size 14px 2 size 14px x size 14px space size 14px sin size 14px space size 14px 3 size 14px x over denominator size 14px minus size 14px tan to the power of size 14px 2 size 14px space size 14px 2 size 14px x end fraction size 14px equals size 14px lim with size 14px x size 14px rightwards arrow size 14px 0 below size 14px minus fraction numerator size 14px 2 size 14px x over denominator size 14px tan size 14px space size 14px 2 size 14px x end fraction size 14px times fraction numerator size 14px sin size 14px space size 14px 3 size 14px x over denominator size 14px tan size 14px space size 14px 2 size 14px x end fraction size 14px equals size 14px minus size 14px 2 over size 14px 2 size 14px times size 14px 3 over size 14px 2 size 14px equals size 14px minus size 14px 3 over size 14px 2 

Maka begin mathsize 14px style limit as x rightwards arrow 0 of space fraction numerator 2 x space sin space 3 x over denominator 1 minus s e c squared space 2 x end fraction equals negative 3 over 2 end style 

0

Roboguru

Nilai x→0lim​tan21​x2sin6x​=....

Pembahasan Soal:

Jawaban: B

2

Roboguru

Jikax→0lim​xsin​=1,makax→0lim​(x22​−x2tanxsin2x​)=...

Pembahasan Soal:

limit as x rightwards arrow 0 of open parentheses 2 over x squared minus fraction numerator sin 2 x over denominator x squared tan x end fraction close parentheses equals...  limit as x rightwards arrow 0 of open parentheses 2 over x squared minus fraction numerator sin 2 x over denominator x squared tan x end fraction close parentheses equals limit as x rightwards arrow 0 of open parentheses 2 over x squared minus fraction numerator sin 2 x over denominator x squared begin display style fraction numerator sin x over denominator cos x end fraction end style end fraction close parentheses  limit as x rightwards arrow 0 of open parentheses 2 over x squared minus fraction numerator sin 2 x over denominator x squared tan x end fraction close parentheses equals limit as x rightwards arrow 0 of open parentheses 2 over x squared minus fraction numerator 2 cos squared x over denominator x squared end fraction close parentheses  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 space equals limit as x rightwards arrow 0 of 2 over x squared left parenthesis 1 minus cos squared x right parenthesis  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 space equals 2 limit as x rightwards arrow 0 of fraction numerator sin squared x over denominator x squared end fraction  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 space equals 2 left parenthesis 1 right parenthesis  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 space equals 2

0

Roboguru

Nilai x→0lim​2xtan2x1−cos24x​=...

Pembahasan Soal:

limit as x rightwards arrow 0 of fraction numerator 1 minus cos squared 4 x over denominator 2 x space tan space 2 x end fraction equals limit as x rightwards arrow 0 of fraction numerator sin squared space 4 x over denominator 2 x space tan space 2 x end fraction equals limit as x rightwards arrow 0 of fraction numerator sin space 4 x space sin space 4 x over denominator 2 x space tan space 2 x end fraction equals fraction numerator 4.4 over denominator 2.2 end fraction equals 4

1

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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