Pertanyaan

Jika L , K adalah bilangan real dan x → c lim f ( x ) = L , x → c lim g ( x ) = K maka tentukan x → c lim ( f ( x ) + g ( x ) f ( x ) − g ( x ) ) 2 !

Jika $L$ , $K$ adalah bilangan real dan $x \to c lim f (x) = L$ , $x \to c lim g (x) = K$ maka tentukan $x \to c lim (\frac{f ( x ) - g ( x )}{f ( x ) + g ( x )})^{2}$ !

F. Ayudhita

Master Teacher

Jawaban terverifikasi

Jawaban

.

$size 14px lim with size 14px x size 14px rightwards arrow size 14px c below begin mathsize 14px style left parenthesis fraction numerator f open parentheses x close parentheses minus g left parenthesis x right parenthesis over denominator f open parentheses x close parentheses plus g left parenthesis x right parenthesis end fraction right parenthesis end style to the power of size 14px 2 size 14px equals fraction numerator begin mathsize 14px style left parenthesis L minus K right parenthesis squared end style over denominator begin mathsize 14px style left parenthesis L plus K right parenthesis squared end style end fraction$ .

Pembahasan

Sifat yang digunakan: 1. ; 2. ; dan 3. . Dapat diperoleh, Jadi, .

Sifat yang digunakan:

1. ;

2. undefined ; dan

3. .

Dapat diperoleh,

$begin mathsize 14px style limit as x rightwards arrow c of open parentheses fraction numerator f open parentheses x close parentheses minus g left parenthesis x right parenthesis over denominator f open parentheses x close parentheses plus g left parenthesis x right parenthesis end fraction close parentheses squared equals open parentheses limit as x rightwards arrow c of fraction numerator f open parentheses x close parentheses minus g left parenthesis x right parenthesis over denominator f open parentheses x close parentheses plus g left parenthesis x right parenthesis end fraction close parentheses squared equals open parentheses fraction numerator limit as x rightwards arrow c of open parentheses f open parentheses x close parentheses minus g left parenthesis x right parenthesis close parentheses over denominator limit as x rightwards arrow c of open parentheses f open parentheses x close parentheses plus g left parenthesis x right parenthesis close parentheses end fraction close parentheses squared equals open parentheses fraction numerator limit as x rightwards arrow c of f left parenthesis x right parenthesis minus limit as x rightwards arrow c of g left parenthesis x right parenthesis over denominator limit as x rightwards arrow c of f left parenthesis x right parenthesis plus limit as x rightwards arrow c of g left parenthesis x right parenthesis end fraction close parentheses squared equals open parentheses fraction numerator L minus K over denominator L plus K end fraction close parentheses squared equals open parentheses L minus K close parentheses squared over open parentheses L plus K close parentheses squared end style$

Jadi, $size 14px lim with size 14px x size 14px rightwards arrow size 14px c below begin mathsize 14px style left parenthesis fraction numerator f open parentheses x close parentheses minus g left parenthesis x right parenthesis over denominator f open parentheses x close parentheses plus g left parenthesis x right parenthesis end fraction right parenthesis end style to the power of size 14px 2 size 14px equals fraction numerator begin mathsize 14px style left parenthesis L minus K right parenthesis squared end style over denominator begin mathsize 14px style left parenthesis L plus K right parenthesis squared end style end fraction$ .

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

849

4.4 (7 rating)

Pertanyaan serupa

Jika L, K adalah bilangan real dan x → 2 lim f ( x ) = L , x → 2 lim g ( x ) = K maka tentukan x → 2 lim ( f ( x ) + g ( x ) f ( x ) − g ( x ) ) 2 .

131

5.0

Jawaban terverifikasi

Nilai dari x → 2 lim 2 x 2 + 3 x + 1 6 x 2 − 7 x − 5 adalah...

5.0

Jawaban terverifikasi

Diketahui x → 2 lim f ( x ) = 3 dan nilai limit berikut. ( i ) x → 2 lim [ f ( x ) − f 2 ( x ) ] = 6 ( ii ) x → 2 lim [ ( f ( x ) + 2 ) 2 − 4 f ( x ) ] = 13 ( iii ) x → 2 lim [ f ( x ) − 2 ...

820

4.2

Jawaban terverifikasi

Tentukan limit fungsi dari: x → 0 lim x − 2 4 x 2 + 1 = ....

0.0

Jawaban terverifikasi

x → 2 lim 12 x + 1 5 x − 1 = . . .

138

5.0

Jawaban terverifikasi