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Nilai dari  adalah ....

Pertanyaan

Nilai dari limit as x rightwards arrow infinity of space open parentheses negative x plus tan space 1 over x close parentheses adalah ....

  1. negative infinity 

  2. 0 

  3. 1 

  4. 2 

  5. infinity 

Pembahasan Soal:

Ingat sifat limit:

limit as x rightwards arrow 0 of space fraction numerator tan space x over denominator x end fraction equals limit as x rightwards arrow 0 of space fraction numerator x over denominator tan space x end fraction equals 1 

MIsalkan y equals 1 over x, maka x equals 1 over y. Untuk x mendekati infinity, maka y mendekati 0.

Penyelesaian dari limit as x rightwards arrow infinity of space open parentheses negative x plus tan space 1 over x close parentheses didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of space open parentheses negative x plus tan space 1 over x close parentheses end cell equals cell limit as y rightwards arrow 0 of space open parentheses negative 1 over y plus tan space y close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of space open parentheses fraction numerator negative 1 plus y space tan space y over denominator y end fraction close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of space open parentheses negative 1 over y plus y times fraction numerator tan space y over denominator y end fraction close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of minus 1 over y plus limit as y rightwards arrow 0 of space y times fraction numerator tan space y over denominator y end fraction end cell row blank equals cell limit as y rightwards arrow 0 of minus 1 over y plus limit as y rightwards arrow 0 of space y times limit as y rightwards arrow 0 of fraction numerator tan space y over denominator y end fraction end cell row blank equals cell negative 1 over 0 plus 0 times 1 end cell row blank equals cell negative infinity plus 0 end cell row blank equals cell negative infinity end cell end table 

Dengan demikian, nilai dari limit as x rightwards arrow infinity of space open parentheses negative x plus tan space 1 over x close parentheses adalah negative infinity.

Jadi, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

E. Lestari

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 11 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai dari  adalah ...

Pembahasan Soal:

Ingat bahwa :

begin mathsize 14px style limit as straight y rightwards arrow 0 of fraction numerator tan space straight y over denominator straight y end fraction equals 1 comma space sec space straight A equals fraction numerator 1 over denominator cos space straight A end fraction comma space straight a space. space straight b equals fraction numerator straight b over denominator begin display style 1 over straight a end style end fraction end style 

Misalkan undefined, sehingga undefined, maka undefined.

Perhatikan bahwa 

begin mathsize 14px style equals limit as straight x rightwards arrow infinity of 2 straight x space tan space open parentheses 1 over straight x close parentheses space. space sec space open parentheses 2 over straight x close parentheses equals limit as straight x rightwards arrow infinity of fraction numerator tan space open parentheses begin display style 1 over straight x end style close parentheses space 2 space sec space open parentheses begin display style 1 over straight x end style close parentheses over denominator begin display style 1 over straight x end style end fraction equals limit as straight x rightwards arrow infinity of fraction numerator tan space straight y space. space 2 space sec space 2 space straight y over denominator straight y end fraction equals limit as straight x rightwards arrow infinity of open parentheses fraction numerator tan space straight y over denominator straight y end fraction space. space 2 space sec space 2 space straight y close parentheses equals limit as straight x rightwards arrow infinity of open parentheses fraction numerator tan space straight y over denominator straight y end fraction space. space fraction numerator 2 over denominator cos space 2 straight y end fraction close parentheses equals 1 space. space fraction numerator 2 over denominator cos space 0 end fraction equals 1 space. space 2 over 1 equals 2 end style 

0

Roboguru

Nilai dari   adalah ....

Pembahasan Soal:

Misalkan begin mathsize 14px style y equals 1 over x end style  , sehingga undefined  . Ketika begin mathsize 14px style x rightwards arrow infinity end style  , maka undefined.

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow straight infinity of invisible function application 3 x tan invisible function application open parentheses 9 over x close parentheses end cell equals cell limit as x rightwards arrow straight infinity of invisible function application 3 x tan invisible function application open parentheses 9 times 1 over x close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application 3 open parentheses 1 over y close parentheses tan invisible function application 9 y end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application fraction numerator 3 tan invisible function application 9 y over denominator y end fraction end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application open parentheses fraction numerator 3 tan invisible function application 9 y over denominator y end fraction times 9 over 9 close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application open parentheses fraction numerator tan invisible function application 9 y over denominator 9 y end fraction times 3 times 9 close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application open parentheses fraction numerator tan invisible function application 9 y over denominator 9 y end fraction times 27 close parentheses end cell row blank equals cell 1 times 27 end cell row blank equals 27 end table end style

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Nilai dari   adalah ....

Pembahasan Soal:

Perhatikan bahwa

  begin mathsize 14px style limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator square root of 2 minus square root of 2 plus 3 over x end root end fraction equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator square root of 2 minus square root of 2 plus 3 over x end root end fraction times fraction numerator square root of 2 plus square root of 2 plus 3 over x end root over denominator square root of 2 plus square root of 2 plus 3 over x end root end fraction close parentheses equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator 2 minus open parentheses 2 plus 3 over x close parentheses end fraction times open parentheses square root of 2 plus square root of 2 plus 3 over x end root close parentheses close parentheses equals limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator negative 3 over x end fraction times open parentheses square root of 2 plus square root of 2 plus 3 over x end root close parentheses close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator negative 3 over x end fraction times limit as x rightwards arrow straight infinity of invisible function application open parentheses square root of 2 plus square root of 2 plus 3 over x end root close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator negative 3 over x end fraction times open parentheses square root of 2 plus square root of 2 plus 0 end root close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator negative 3 over x end fraction times open parentheses 2 square root of 2 close parentheses end style      

Misalkan begin mathsize 14px style y equals 1 over x end style. Ketika begin mathsize 14px style x rightwards arrow infinity end style, maka begin mathsize 14px style y rightwards arrow 0 end style. Sehingga

begin mathsize 14px style limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application open parentheses 6 over x close parentheses over denominator negative 3 over x end fraction times open parentheses 2 square root of 2 close parentheses equals limit as x rightwards arrow straight infinity of invisible function application fraction numerator tan invisible function application 6 open parentheses 1 over x close parentheses over denominator negative 3 open parentheses 1 over x close parentheses end fraction times open parentheses 2 square root of 2 close parentheses equals limit as y rightwards arrow 0 of invisible function application fraction numerator tan invisible function application 6 y over denominator negative 3 y end fraction times open parentheses 2 square root of 2 close parentheses equals limit as y rightwards arrow 0 of invisible function application fraction numerator tan invisible function application 6 y over denominator negative 3 y end fraction times 2 over 2 times open parentheses 2 square root of 2 close parentheses equals limit as y rightwards arrow 0 of invisible function application fraction numerator tan invisible function application 6 y over denominator 6 y end fraction times fraction numerator 2 over denominator negative 1 end fraction times open parentheses 2 square root of 2 close parentheses equals limit as y rightwards arrow 0 of invisible function application fraction numerator tan invisible function application 6 y over denominator 6 y end fraction times open parentheses negative 4 square root of 2 close parentheses equals 1 times open parentheses negative 4 square root of 2 close parentheses equals negative 4 square root of 2 end style

Jadi, jawaban yang tepat adalah E.

0

Roboguru

Nilai dari  adalah ....

Pembahasan Soal:

Misalkan begin mathsize 14px style y space equals space 1 over x end style  ,sehingga begin mathsize 14px style x space equals space 1 over y end style  . Ketika begin mathsize 14px style x rightwards arrow infinity end style  , maka begin mathsize 14px style space y rightwards arrow 0 end style  .

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow straight infinity of invisible function application x tan invisible function application open parentheses 1 over x close parentheses end cell equals cell limit as y rightwards arrow 0 of invisible function application 1 over y tan invisible function application y end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application fraction numerator tan invisible function application y over denominator y end fraction end cell row blank equals 1 end table end style 

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Nilai dari  adalah ...

Pembahasan Soal:

Misalkan begin mathsize 14px style 1 over straight x equals straight y end style, sehingga begin mathsize 14px style cot space straight y equals fraction numerator 1 over denominator tan space straight y end fraction end style. Ketika begin mathsize 14px style straight x rightwards arrow infinity end style, maka begin mathsize 14px style straight y rightwards arrow 0 end style.

Maka

begin mathsize 14px style equals limit as straight x rightwards arrow infinity of fraction numerator 2 straight x space cot space open parentheses begin display style 2 over straight x end style close parentheses minus 3 space cot space open parentheses begin display style 2 over straight x end style close parentheses over denominator 5 straight x squared minus 2 straight x end fraction equals limit as straight x rightwards arrow infinity of fraction numerator open parentheses 2 straight x minus 3 close parentheses space cot space open parentheses begin display style 2 over straight x end style close parentheses over denominator straight x space open parentheses 5 straight x minus 2 close parentheses end fraction equals limit as straight x rightwards arrow infinity of fraction numerator open parentheses 2 straight x minus 3 close parentheses over denominator open parentheses 5 straight x minus 2 close parentheses end fraction space. space 1 over straight x space. space cot space open parentheses 2 over straight x close parentheses equals limit as straight x rightwards arrow infinity of fraction numerator open parentheses 2 straight x minus 3 close parentheses over denominator open parentheses 5 straight x minus 2 close parentheses end fraction space. space limit as straight x rightwards arrow infinity of 1 over straight x space. space cot space open parentheses 2 over straight x close parentheses equals 2 over 5 space. space limit as straight y rightwards arrow 0 of straight y space. space cot space 2 straight y equals 2 over 5 space. space limit as straight y rightwards arrow 0 of straight y space. space fraction numerator 1 over denominator tan space 2 straight y end fraction equals 2 over 5 space. space limit as straight y rightwards arrow 0 of space fraction numerator straight y over denominator tan space 2 straight y end fraction equals 2 over 5 space. space 1 half equals 1 fifth end style   

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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