YA
Yesi A
30 Juli 2024 12:47
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YA
Yesi A
30 Juli 2024 12:47
Pertanyaan
lim x -> 0 (cos x-1)/(tan² x)
lim x -> 0 (cos x-1)/(tan² x)
1
1
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AG
Adley G
30 Juli 2024 12:49
Jawaban terverifikasi
<p>Lim ( (cos x - 1) / (tan² 2x))<br><br>x→0<br><br>= lim ( (cos x - 1) / tan² 2x ) . ( (cos x + 1) / (cos x + 1) )<br><br>...x→0<br><br>= lim ( ( cos² x - 1 ) / tan² 2x ) . 1/( cos x + 1 )<br><br>...x→0<br><br>= lim ( (1 - sin² x - 1) / tan² 2x) . 1/(cos x + 1)<br><br>...x→0<br><br>= lim (( -sin² x) / tan² 2x) . (1/(cos x + 1))<br><br>...x→0<br><br>= (-(1)²/ (2)²) . (1/(cos 0 + 1))<br><br>= - ( 1/4 ) . ( 1/(1 + 1) )<br><br>= - 1/4 . ( 1/2 )<br><br>= - 1/8</p>
Lim ( (cos x - 1) / (tan² 2x))
x→0
= lim ( (cos x - 1) / tan² 2x ) . ( (cos x + 1) / (cos x + 1) )
...x→0
= lim ( ( cos² x - 1 ) / tan² 2x ) . 1/( cos x + 1 )
...x→0
= lim ( (1 - sin² x - 1) / tan² 2x) . 1/(cos x + 1)
...x→0
= lim (( -sin² x) / tan² 2x) . (1/(cos x + 1))
...x→0
= (-(1)²/ (2)²) . (1/(cos 0 + 1))
= - ( 1/4 ) . ( 1/(1 + 1) )
= - 1/4 . ( 1/2 )
= - 1/8
· 0.0 (0)
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