✔ 最佳答案
Note: This should be a 1-sided limit, since the domain of y = (sin(4x))^(tan(3x)) can not include negative real numbers...
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Use logarithms.
Let L = lim(x→0+) (sin(4x))^(tan(3x)).
So, ln L = lim(x→0+) tan(3x) ln(sin(4x))
.............= lim(x→0+) ln(sin(4x)) / cot(3x); now of the form ∞/∞
.............= lim(x→0+) (4 cos(4x)/sin(4x)) / (-3 csc^2(3x)); by L'Hopital's Rule
.............= (-4/3) * lim(x→0+) cos(4x) * sin^2(3x)/sin(4x)
.............= (-4/3) * lim(x→0+) cos(4x) * x^2 sin^2(3x)/(x^2 sin(4x))
.............= (-4/3) * lim(x→0+) x cos(4x) * (sin(3x) / x)^2 * (x/sin(4x))
.............= (-4/3) * lim(x→0+) x cos(4x) * (3 sin(3x) / (3x))^2 * (4x/(4 sin(4x)))
.............= (-1/3) * lim(x→0+) x cos(4x) * 9(sin(3x) / (3x))^2 * (4x/sin(4x))
.............= (-1/3) * 0 * 1 * 9 * 1^2 * 1/1, via lim(t→0) sin(t)/t = 1
.............= 0.
Hence, L = e^0 = 1.
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I hope this helps!
