Limits with variable substitution and Trig \lim_{x\to0}\frac{\sin^2(3x)}{x^2\cos x}=\lim_{u\to0}\frac{\sin^2(u)}{(\frac u3)^2}=9(\lim_{u\to0}\frac{\sin u}u)^2

albusgks 2022-04-29 Answered
Limits with variable substitution and Trig
limx0sin2(3x)x2cosx=limu0sin2(u)(u3)2=9(limu0sinuu)2
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Answers (1)

timbreoizy
Answered 2022-04-30 Author has 15 answers
Hint:
limx0sin2(3x)x2cosx=
limx0sin(3x)x.limx0sin(3x)x.limx01cosx=
limx03sin(3x)3x.limx03sin(3x)3x.limx01cosx=
9limx0sin(3x)3x1.limx0sin(3x)3x1.limx01cosx=
9.1.1.limx01cosx=9.1.1.1
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