Evaluating \(\displaystyle\lim_{{{x}\to{0}}}{\frac{{{x}}}{{{\sin{{\left(\frac{{1}}{{x}^{{2}}}\right)}}}}}}\)

Harold Hoover

Harold Hoover

Answered question

2022-04-05

Evaluating limx0xsin(1x2)

Answer & Explanation

kanonickiuoeh

kanonickiuoeh

Beginner2022-04-06Added 8 answers

Let consider for n+
1xn2=1n+2nπ+xn=11n+2nπ0sin(1xn2)=1n+o(1n)
1yn2=1n+2nπ+yn=11n+2nπ0sin(1yn2)=1n+o(1n)
thus
xnsin(1xn2)=11n+2nπ(1n+o(1n))=nn1+2n2π(1+o(1))+
ynsin(1yn2)=11n+2nπ(1n+o(1n))=nn1+2n2π(1+o(1))
then the limit does not exist.

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