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pakai deret maclaurin biar gampang.
misal, f1(x) = cos(5x/3)
f1(0) = 1
f1'(x) = -(5/3)sin(5x/3)
f1'(0) = 0
f1''(x) = -(25/9)cos(5x/3)
f1''(0) = -(25/9)
dst...
= f(a) + (x - a)/1! * f'(a) + (x - a)^2/2! * f''(a) + ...
= 1 + (x - 0)*0 + (x - 0)^2/2 * (-25/9) + ...
= 1 - (25/9)x^2 + ...
maka,
1 - cos(5x/3) = 1 - (1 - (25/9)x^2 + ...)
= (25/9)x^2 + (625/1944)x^4 + ...
f2(x) = (3/4x)^2 sin(x/2) cot(2x)
f2(0) = 0
f2'(x) = (9/32)x^2 cos(x/2)cot(2x) - (9/8)x^2 sin(2/x)csc^2(2x) + (9/8)xsin(x/2)cot(2x)
f2'(x) = 0
dan seterusnya, sehingga didapatkan deret :
= 9x^2/64 - 99x^4/512 + ...
Jadi, nilainya adalah
(25x^2/18 - 625x^4/1944 + ... ) / (9x^2/64 - 99x^4/512 + ...)
Karena limit x mendekati 0, kita gunakan yang depannya saja (karena ketika dibagi, suku suku berikutnya akan ada nilai x yang jika disubtitusi dengan 0 hasilnya akan 0 )
(25/18) / (9/64)
= (25/18) * (64/9)
= 1600/162
= 800/81
CMIIW