Question

Materi Limit, kerjakan nomor 7 dan 8. Terimakasih

Answer 1

Limit mendekati 0

x + sin2x / x - sin2x
bagi dengan x
((x + sin2x)/x) / ((x - sin2x)/x)
(x/x + sin2x/x) / (x/x - sin2x/x)
= 1 + sin(2x/x) / 1 - sin(2x/x)
= 1 + 2 * 1 / 1 - sin 2 * 1
= 1 + 2 / 1 - 2
= 3 / -1
= -3

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lim x mendekati π

sinx / √x - π

U = sinx
U' = cosx

V = √x - π = (x - π)^1/2
V' = 1/2(x-π)^-1/2
V' = 1 / 2√x-π

limit x mendekati π

cosx / (1 / 2√x-π)
cos90 / (1 / 2√0)
= 0 / 0
= Tak terdefinisi