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Nilai x→0lim​5−x2+9​4−x​=…

Pertanyaan

Nilai begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator 4 minus x over denominator 5 minus square root of x squared plus 9 end root end fraction equals horizontal ellipsis end style

  1. begin mathsize 14px style 2 end style

  2. begin mathsize 14px style 5 over 3 end style

  3. begin mathsize 14px style 10 over 7 end style

  4. begin mathsize 14px style 5 over 4 end style

  5. begin mathsize 14px style 10 over 9 end style

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator 5 minus square root of x squared plus 9 end root end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator 4 minus x over denominator 5 minus square root of x squared plus 9 end root end fraction cross times fraction numerator 5 plus square root of x squared plus 9 end root over denominator 5 plus square root of x squared plus 9 end root end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator open parentheses 4 minus x close parentheses. open parentheses 5 plus square root of x squared plus 9 end root close parentheses over denominator 25 minus open parentheses x squared plus 9 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator open parentheses 4 minus x close parentheses. open parentheses 5 plus square root of x squared plus 9 end root close parentheses over denominator 16 minus x squared end fraction end cell row blank equals cell limit as x rightwards arrow 4 of fraction numerator open parentheses 4 minus x close parentheses. open parentheses 5 plus square root of x squared plus 9 end root close parentheses over denominator open parentheses 4 minus x close parentheses. open parentheses 4 plus x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of invisible function application fraction numerator open parentheses 5 plus square root of x squared plus 9 end root close parentheses over denominator open parentheses 4 plus x close parentheses end fraction end cell row blank equals cell fraction numerator 5 plus square root of 4 squared plus 9 end root over denominator 4 plus 4 end fraction end cell row blank equals cell 10 over 8 end cell row blank equals cell 5 over 4 end cell end table end style

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Nilai dari x→−2lim​2x+48x2+14x−4​=….

Pembahasan Soal:

Nilai suatu limit dapat ditentukan dengan beberapa cara, salah satunya dengan menggunakan metode pemfaktoran, hal ini dilakukan saat metode substitusi menghasilkan nilai limit undefined.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow negative 2 of fraction numerator left parenthesis 2 x plus 4 right parenthesis left parenthesis 4 x minus 1 right parenthesis over denominator left parenthesis 2 x plus 4 right parenthesis end fraction end cell equals cell limit as straight x rightwards arrow negative 2 of left parenthesis 4 x minus 1 right parenthesis end cell row blank equals cell 4 open parentheses negative 2 close parentheses minus 1 end cell row blank equals cell negative 8 minus 1 end cell row blank equals cell negative 9 end cell end table end style

Jadi, jawaban yang tepat adalah A.

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Roboguru

x→1lim​2+2x​−6−2x​x3−x2​=

Pembahasan Soal:

undefined

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Roboguru

Hasil dari x→2lim​1+x3x2−4​ adalah ....

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator x squared minus 4 over denominator 1 plus x cubed end fraction end cell equals cell fraction numerator 2 squared minus 4 over denominator 1 plus 2 cubed end fraction end cell row blank equals cell fraction numerator 4 minus 4 over denominator 1 plus 8 end fraction end cell row blank equals cell 0 over 9 end cell row blank equals 0 end table end style 

Jadi, jawaban yang tepat adalah C.undefined 

0

Roboguru

x→−1lim​3x​+1x+1​=...

Pembahasan Soal:

limit as straight x rightwards arrow negative 1 of fraction numerator straight x plus 1 over denominator cube root of straight x plus 1 end fraction equals...

Penyelesaian 

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow negative 1 of fraction numerator straight x plus 1 over denominator cube root of straight x plus 1 end fraction end cell equals cell... end cell row blank equals cell limit as straight x rightwards arrow negative 1 of fraction numerator straight x plus 1 over denominator cube root of straight x plus 1 end fraction straight x fraction numerator cube root of straight x minus 1 over denominator cube root of straight x minus 1 end fraction end cell row blank equals cell limit as straight x rightwards arrow negative 1 of fraction numerator open parentheses straight x plus 1 close parentheses cube root of straight x minus 1 over denominator straight x to the power of begin display style 2 end style end exponent minus 1 end fraction end cell row blank equals cell limit as straight x rightwards arrow negative 1 of fraction numerator open parentheses x plus 1 close parentheses cube root of x minus 1 over denominator open parentheses x minus 1 close parentheses open parentheses x plus 1 close parentheses end fraction end cell row blank equals cell limit as straight x rightwards arrow negative 1 of fraction numerator cube root of x minus 1 over denominator x minus 1 end fraction end cell row blank equals cell fraction numerator negative 1 minus 1 over denominator negative 1 minus 1 end fraction end cell row blank equals 1 end table

Jadi nilai dari limit as x rightwards arrow negative 1 of fraction numerator x plus 1 over denominator cube root of x plus 1 end fraction equals 1

Oleh karena itu, jawaban yang benar adalah B.

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Roboguru

Nilai x→−1lim​x−1x2−1​=…

Pembahasan Soal:

Ingat cara menentukan nilai sebuah limit dengan cara pemfaktoran dari persamaan yang ada.

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

Jadi, dapat disimpulkan bahwa nilai dari limit as x rightwards arrow negative 1 of invisible function application fraction numerator x squared minus 1 over denominator x minus 1 end fraction adalah 0.

Oleh karena itu, jawaban yang benar adalah D.

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