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Diketahui  dan . Tentukan :

Pertanyaan

Diketahui limit as straight x rightwards arrow straight a of invisible function application straight f left parenthesis straight x right parenthesis equals 8 comma blank limit as straight x rightwards arrow straight a of invisible function application straight g left parenthesis straight x right parenthesis equals negative 3 dan limit as straight x rightwards arrow straight a of invisible function application straight h left parenthesis straight x right parenthesis equals 4. Tentukan :

limit as straight x rightwards arrow straight a of invisible function application fraction numerator straight h open parentheses straight x close parentheses over denominator straight f left parenthesis straight x right parenthesis end fraction open parentheses straight f open parentheses straight x close parentheses times straight g open parentheses straight x close parentheses plus 2 straight h left parenthesis straight x right parenthesis close parentheses

Pembahasan Soal:

Dalam menyelesaikan soal di atas, ingat sifat limit di bawah ini.

left parenthesis straight i right parenthesis space limit as straight x rightwards arrow straight c of invisible function application kf left parenthesis straight x right parenthesis equals straight k limit as straight x rightwards arrow straight c of invisible function application straight f left parenthesis straight x right parenthesis left parenthesis ii right parenthesis space limit as straight x rightwards arrow straight c of invisible function application open square brackets straight f left parenthesis straight x right parenthesis times straight g left parenthesis straight x right parenthesis close square brackets equals open square brackets limit as straight x rightwards arrow straight c of invisible function application straight f left parenthesis straight x right parenthesis close square brackets open square brackets limit as straight x rightwards arrow straight c of invisible function application straight g left parenthesis straight x right parenthesis close square brackets left parenthesis iii right parenthesis space space limit as straight x rightwards arrow straight c of invisible function application open square brackets straight f open parentheses straight x close parentheses plus-or-minus straight g left parenthesis straight x right parenthesis close square brackets equals limit as straight x rightwards arrow straight c of invisible function application straight f left parenthesis straight x right parenthesis plus-or-minus limit as straight x rightwards arrow straight c of invisible function application straight g left parenthesis straight x right parenthesis left parenthesis iv right parenthesis space limit as straight x rightwards arrow straight c of invisible function application open square brackets fraction numerator straight f open parentheses straight x close parentheses over denominator straight g left parenthesis straight x right parenthesis end fraction close square brackets equals fraction numerator limit as straight x rightwards arrow straight c of invisible function application straight f open parentheses straight x close parentheses over denominator limit as straight x rightwards arrow straight c of invisible function application straight g open parentheses straight x close parentheses end fraction left parenthesis straight v right parenthesis space limit as straight x rightwards arrow straight c of invisible function application open square brackets straight f left parenthesis straight x right parenthesis close square brackets to the power of straight n equals open square brackets limit as straight x rightwards arrow straight c of invisible function application straight f left parenthesis straight x right parenthesis close square brackets to the power of straight n    
 

Pada soal, diketahui limit as straight x rightwards arrow straight a of invisible function application straight f left parenthesis straight x right parenthesis equals 8 comma blank limit as straight x rightwards arrow straight a of invisible function application straight g left parenthesis straight x right parenthesis equals negative 3, dan limit as straight x rightwards arrow straight a of invisible function application straight h left parenthesis straight x right parenthesis equals 4. Maka penyelesaian soal di atas adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow straight a of invisible function application fraction numerator straight h open parentheses straight x close parentheses over denominator straight f left parenthesis straight x right parenthesis end fraction open parentheses straight f open parentheses straight x close parentheses times straight g open parentheses straight x close parentheses plus 2 straight h left parenthesis straight x right parenthesis close parentheses end cell equals cell limit as straight x rightwards arrow straight a of invisible function application open parentheses straight g open parentheses straight x close parentheses times straight h open parentheses straight x close parentheses plus 2 fraction numerator open square brackets straight h left parenthesis straight x right parenthesis close square brackets squared over denominator straight f left parenthesis straight x right parenthesis end fraction close parentheses end cell row blank equals cell limit as straight x rightwards arrow straight a of invisible function application open square brackets straight g left parenthesis straight x right parenthesis times straight h left parenthesis straight x right parenthesis close square brackets plus 2 limit as straight x rightwards arrow straight a of invisible function application open square brackets fraction numerator open square brackets straight h left parenthesis straight x right parenthesis close square brackets squared over denominator straight f left parenthesis straight x right parenthesis end fraction close square brackets end cell row blank equals cell open square brackets limit as straight x rightwards arrow straight a of invisible function application straight g left parenthesis straight x right parenthesis close square brackets times open square brackets limit as straight x rightwards arrow straight a of invisible function application straight h left parenthesis straight x right parenthesis close square brackets plus 2 fraction numerator open square brackets limit as straight x rightwards arrow straight a of invisible function application straight h left parenthesis straight x right parenthesis close square brackets squared over denominator limit as straight x rightwards arrow straight a of invisible function application straight f left parenthesis straight x right parenthesis end fraction end cell row blank equals cell open parentheses negative 3 close parentheses 4 plus 2 open parentheses 4 squared over 8 close parentheses end cell row blank equals cell negative 12 plus 2 open parentheses 16 over 8 close parentheses end cell row blank equals cell negative 12 plus 2 open parentheses 2 close parentheses end cell row blank equals cell negative 12 plus 4 end cell row blank equals cell negative 8 end cell end table  


Jadi, nilai limit dari =limit as straight x rightwards arrow straight a of invisible function application fraction numerator straight h open parentheses straight x close parentheses over denominator straight f left parenthesis straight x right parenthesis end fraction open parentheses straight f open parentheses straight x close parentheses times straight g open parentheses straight x close parentheses plus 2 straight h left parenthesis straight x right parenthesis close parentheses yang nilai limit masing-masing fungsinya diketahui adalah negative 8.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Nuryani

Mahasiswa/Alumni Universitas Padjadjaran

Terakhir diupdate 01 Mei 2021

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Pertanyaan yang serupa

Diberikan x→alim​f(x)=3 dan .  Hitunglah nilai setiap limit berikut. a.   b.   c.   d.

Pembahasan Soal:

Ingat bahwa:

Limit memiliki sifat seperti di bawah ini.

1. xalim[f(x)+g(x)]=xalimf(x)+xalimg(x)

2. xalim[f(x)g(x)]=xalimf(x)xalimg(x)

3. xalim[f(x)×g(x)]=xalimf(x)×xalimg(x)

4. xalimkf(x)=k×xalimf(x)

5. xalim[g(x)f(x)]=limxag(x)limxaf(x)

6. xalim[f(x)]n=[xalimf(x)]n

7. xalimf(x)=xalimf(x)

 

Diberikan xalimf(x)=3danxalimg(x)=1

a. xalimf2(x)+g2(x)

Berdasarkan sifat 1, 6, dan 7 diperoleh:

limxaf2(x)+g2(x)====[limxaf(x)]2+[limxag(x)]2[3]2+[1]29+110

Dengan demikian, hasil dari xalimf2(x)+g2(x) adalah 10.

b. xalimf(x)+g(x)2f(x)3g(x)

Berdasarkan sifat 1, 2, 4, dan 5 diperoleh:

limxaf(x)+g(x)2f(x)3g(x)====limxaf(x)+limxag(x)2limxaf(x)3limxag(x)3+(1)2[3]3[1]26+329

Dengan demikian, hasil dari xalimf(x)+g(x)2f(x)3g(x) adalah 29.

c. xalim3g(x)[f(x)+3]

Berdasarkan sifat 1, 3, dan 7 diperoleh:

limxa3g(x)[f(x)+3]====3limxag(x)×limxaf(x)+331×[3+3]1×66

Dengan demikian, hasil dari xalim3g(x)[f(x)+3] adalah 6.

d. xalim[f(x)3]4

Berdasarkan sifat 1 dan 6 diperoleh:

limxa[f(x)3]4=====[limxa[f(x)3]]4[limxaf(x)limxa3]4[33]4[0]40

Dengan demikian, hasil dari xalim[f(x)3]4 adalah 0.

Roboguru

Diketahui  dan . Nilai limit berikut yang benar adalah...

Pembahasan Soal:

Dengan menggunakan konsep dasar limit dan sifat-sifat limit diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow a of left square bracket f open parentheses x close parentheses squared minus g open parentheses x close parentheses squared right square bracket end cell equals cell limit as x rightwards arrow a of f open parentheses x close parentheses squared minus limit as x rightwards arrow a of g open parentheses x close parentheses squared end cell row blank equals cell open parentheses limit as x rightwards arrow a of f left parenthesis x right parenthesis close parentheses squared minus space open parentheses limit as x rightwards arrow a of g open parentheses x close parentheses close parentheses squared end cell row blank equals cell 12 squared minus 2 squared end cell row blank equals cell 144 minus 4 end cell row blank equals 140 end table end style

Oleh karena itu, jawaban yang benar adalah E.

Roboguru

Diketahui x→alim​f(x)=3, , dan . Hitunglah nilai setiap limit berikut a. x→alim​[2f(x)+g(x)]2   b. x→alim​4g(x)f(x)⋅h(x)​  c. x→alim​{f3(x)−2g(x)+5h(x)}

Pembahasan Soal:

Diketahui xalimf(x)=3,  begin mathsize 14px style limit as x rightwards arrow a of h left parenthesis x right parenthesis equals 0 end style dan begin mathsize 14px style limit as x rightwards arrow a of g left parenthesis x right parenthesis equals 5 end style

Ingat 

xalim[f(x)+g(x)]=xalimf(x)+xalimg(x)xalim[f(x)g(x)]=xalimf(x)xalimg(x)xalim[g(x)f(x)]=limxag(x)limxaf(x),xalimg(x)=0 

Perhatikan perhitungan berikut

a. xalim[2f(x)+g(x)]2

 ========limxa[2f(x)+g(x)]2limxa[4f2(x)+4f(x)g(x)+g2(x)]limxa4f2(x)+limxa(4f(x)g(x))+limxag2(x)4f2(a)+4limxaf(x)limxag(x)+524f2(a)+4f(a)5+254f2(a)+20f(a)+254(32)+20(3)+254(9)+60+25121   

Dengan demikian, nilai dari xalim[2f(x)+g(x)]2 adalah 121.

b. xalim4g(x)f(x)h(x) 

limxa4g(x)f(x)h(x)===4limxag(x)limxaf(x)limxah(x)45300  

Dengan demikian, nilai dari xalim4g(x)f(x)h(x) adalah 0.

c. xalim{f3(x)2g(x)+5h(x)}

====limxa{f3(x)2g(x)+5h(x)}limxaf3(x)2limxag(x)+5limxah(x)3325+50271017  

Dengan demikian, nilai dari xalim{f3(x)2g(x)+5h(x)} adalah 17.

Roboguru

Tentukan nilai

Pembahasan Soal:

Dengan menggunakan konsep limit diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator x cubed plus 5 x squared plus 12 x over denominator x cubed plus 8 end fraction end cell equals cell fraction numerator 2 cubed plus 5 open parentheses 2 close parentheses squared plus 12 open parentheses 2 close parentheses over denominator 2 cubed plus 8 end fraction end cell row blank equals cell fraction numerator 8 plus 20 plus 24 over denominator 8 plus 8 end fraction end cell row blank equals cell 52 over 16 end cell row blank equals cell 13 over 4 end cell end table

Dengan demikian nilai 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 cell fraction numerator x cubed plus 5 x squared plus 12 x over denominator x cubed plus 8 end fraction end cell end table adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 13 over 4 end cell end table.

Roboguru

Nilai dari  adalah...

Pembahasan Soal:

Perhatikan perhitungan berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of space fraction numerator 6 x squared minus 7 x minus 5 over denominator 2 x squared plus 3 x plus 1 end fraction end cell equals cell space fraction numerator 6 open parentheses 2 close parentheses squared minus 7 open parentheses 2 close parentheses minus 5 over denominator 2 open parentheses 2 close parentheses squared plus 3 open parentheses 2 close parentheses plus 1 end fraction end cell row blank equals cell fraction numerator 24 minus 14 minus 5 over denominator 8 plus 6 plus 1 end fraction end cell row blank equals cell 5 over 15 end cell row blank equals cell 1 third end cell end table

Dengan demikian nilai dari limit as x rightwards arrow 2 of space fraction numerator 6 x squared minus 7 x minus 5 over denominator 2 x squared plus 3 x plus 1 end fraction adalah 1 third.

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