Limits with variable substitution and Trig \(\displaystyle\lim_{{{x}\to{0}}}{\frac{{{{\sin}^{{2}}{\left({3}{x}\right)}}}}{{{x}^{{2}}{\cos{{x}}}}}}=\lim_{{{u}\to{0}}}{\frac{{{{\sin}^{{2}}{\left({u}\right)}}}}{{{\left({\frac{{u}}{{3}}}\right)}^{{2}}}}}={9}{\left(\lim_{{{u}\to{0}}}{\frac{{{\sin{{u}}}}}{{u}}}\right)}^{{2}}\)

kembdumatxf 2022-03-23 Answered
Limits with variable substitution and Trig
limx0sin2(3x)x2cosx=limu0sin2(u)(u3)2=9(limu0sinuu)2
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Answers (1)

Riya Erickson
Answered 2022-03-24 Author has 12 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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