Find: \(\displaystyle{L}=\lim_{{{x}\to{0}}}{\frac{{{\sin{{\left({1}-{\frac{{{\sin{{\left({x}\right)}}}}}{{{x}}}}\right)}}}}}{{{x}^{{2}}}}}\)

Kale Bright

Kale Bright

Answered question

2022-04-08

Find:
L=limx0sin(1sin(x)x)x2

Answer & Explanation

ysnlm8eut

ysnlm8eut

Beginner2022-04-09Added 14 answers

In a slightly different way, using the Taylor expansion, as x0
sinx=xx36+O(x5)
gives
1sinxx=x26+O(x4)
then
sin(1sinxx)=x26+O(x4)
and
sin(1sinxx)x2=16+O(x2)
from which one may conclude easily.

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