Roboguru

h→0lim​hx+h​−x​​

Pertanyaan

begin mathsize 14px style limit as h rightwards arrow 0 of fraction numerator square root of x plus h end root minus square root of x over denominator h end fraction end style

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as h rightwards arrow 0 of fraction numerator square root of x plus h end root minus square root of x over denominator h end fraction end cell row blank equals cell limit as h rightwards arrow 0 of fraction numerator square root of x plus h end root minus square root of x over denominator h end fraction cross times fraction numerator square root of x plus h end root plus square root of x over denominator square root of x plus h end root plus square root of x end fraction end cell row blank equals cell limit as h rightwards arrow 0 of fraction numerator x plus h minus x over denominator h open parentheses square root of x plus h end root plus square root of x close parentheses end fraction end cell row blank equals cell limit as h rightwards arrow 0 of fraction numerator up diagonal strike h over denominator up diagonal strike h open parentheses square root of x plus h end root plus square root of x close parentheses end fraction end cell row blank equals cell limit as h rightwards arrow 0 of fraction numerator 1 over denominator square root of x plus h end root plus square root of x end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of x plus 0 end root plus square root of x end fraction end cell row blank equals cell fraction numerator 1 over denominator 2 square root of x end fraction end cell end table end style

Jadi, jawabannya adalah begin mathsize 14px style fraction numerator bold 1 over denominator bold 2 square root of bold x end fraction end style

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Nilai dari x→−2lim​x−2x2+7x+2​=....

Pembahasan Soal:

Dengan cara substitusi, maka didapatkan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 of fraction numerator x squared plus 7 x plus 2 over denominator x minus 2 end fraction end cell equals cell fraction numerator open parentheses negative 2 close parentheses squared plus 7 left parenthesis negative 2 right parenthesis plus 2 over denominator negative 2 minus 2 end fraction end cell row blank equals cell fraction numerator 4 minus 14 plus 2 over denominator negative 4 end fraction end cell row blank equals cell fraction numerator negative 8 over denominator negative 4 end fraction end cell row blank equals 2 end table end style

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Jika x→3lim​f(x)=5 dan x→3lim​g(x)=9, maka nilai x→3lim​2f(x)f(x)+g(x)​=...

Pembahasan Soal:

Perhatikan perhitungan berikut.

Ingat:

begin mathsize 14px style limit as x rightwards arrow c of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equals fraction numerator limit as x rightwards arrow c of f open parentheses x close parentheses over denominator limit as x rightwards arrow c of g open parentheses x close parentheses end fraction end style

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

Maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator f open parentheses x close parentheses plus g open parentheses x close parentheses over denominator 2 f open parentheses x close parentheses end fraction end cell equals cell fraction numerator 5 plus 9 over denominator 2 times 5 end fraction end cell row blank equals cell 14 over 10 end cell row blank equals cell 1 comma 4 end cell end table end style

Jadi, nilai limit tersebut adalah mendekati begin mathsize 14px style 1 comma 4 end style.

Dengan demikian, jawaban yang tepat adalah C.

1

Roboguru

Find the value of the limit and which applicable limit theorems used to find the value.  b. r→1lim​r+38r+1​​

Pembahasan Soal:

0

Roboguru

x→4lim​2−x​4−x​

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator 2 minus square root of x end fraction end cell equals cell limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator 2 minus square root of x end fraction cross times fraction numerator 2 plus square root of x over denominator 2 plus square root of x end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator up diagonal strike open parentheses 4 minus x close parentheses end strike. open parentheses 2 plus square root of x close parentheses over denominator up diagonal strike 4 minus x end strike end fraction end cell row blank equals cell limit as x rightwards arrow 4 of 2 plus square root of x end cell row blank equals cell 2 plus square root of 4 end cell row blank equals 4 end table end style

Jadi, jawabannya adalah 4.

0

Roboguru

x→5lim​x2−3x−10x2−10x+25​

Pembahasan Soal:

Dengan menggunakan metode substitusi didapatkan

limx5x23x10x210x+25===523(5)105210(5)+252515102550+2500

Sehingga perlu digunakan metode lain untuk menentukan nilai limit, yaitu dengan cara pemfaktoran sebagai berikut

limx5x23x10x210x+25=====limx5(x5)(x+2)(x5)(x5)limx5(x+2)(x5)5+255700

Dengan demikian, nilai dari limit as x rightwards arrow 5 of fraction numerator x squared minus 10 x plus 25 over denominator x squared minus 3 x minus 10 end fraction equals 0.

0

Roboguru

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