Question

Mohon bantuannya no II-III(a,b,c,) sangat butuh cara jalannya dan please jaangan jawab asal:"
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Answer 1

Verified answer

Ralat :
no. 1
lim [x-->0] sin² 2x /tan³ 5x
= lim [x-->0] 1/tan 5x . lim [x-->0] sin² 2x/tan² 3x
= lim [x-->0] 1/tan 5x . lim [x-->0] (2x/5x)²
= (1/tan 0) . (2/5)²
= ~

No. 2
lim [x-->0] (1 - cos x)/3x²
= lim [x-->0] 2sin² (1/2 x) /3x²
= lim [x-->0] 2sin² (1/2 x) . (1/2)² /(3 . (1/2)² x²)
= (2/4)/3
= 1/6

Jawaban pd lampiran...