Evaluate: limx→0sin3x−x3 sgn(1−[xsin−1x])xtan2xsin⁡(πcos⁡x)(Note that here sgn denotes the Signum function and [x] denotes the greatest integer less than or equal to x.)The above simplifies to:limx→0sin3(x)−x3xtan2(x)sin⁡(π(1−cos⁡x))

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Evaluate: limx→0sin3x−x3 sgn(1−[xsin−1x])xtan2xsin⁡(πcos⁡x)(Note that here sgn denotes the Signum function and [x] denotes the greatest integer less than or equal to x.)The above simplifies to:limx→0sin3(x)−x3xtan2(x)sin⁡(π(1−cos⁡x))

Jermaine Lam · Accepted Answer

We have thatsin3x−x3xtan2xsin⁡(π(1−cos⁡x))=x3−12x5+o(x5)−x3x3(x2+o(x2))π(1−cos⁡x)x2sin⁡(π(1−cos⁡x))π(1−cos⁡x)=-12x5+o(x5)(x5+o(x5))π(1-cosx)x2sin(π(1-cosx))π(1-cosx)=-12+o(1)(1+o(1))π(1-cosx)x2sin(π(1-cosx))π(1-cosx)→-121·π2·1=-1πWithout Taylor&#039;s expansion note that we havesin3x−x3xtan2xsin⁡(π(1−cos⁡x))=sin3x−x3x51tan2xx2π(1−cos⁡x)x2sin⁡(π(1−cos⁡x))π(1−cos⁡x)then since by standard limits the second part&amp;nbsp;→2π&amp;nbsp;we can analizesin3x−x3x5=(sin⁡x−x)(sin2x+sin⁡x⋅x+x2)x5=sin⁡x−xx3sin2x+sin⁡x⋅x+x2x2and since by standard limits&amp;nbsp;sin2x+sin⁡x⋅x+x2x2→3&amp;nbsp;we need to considersin⁡x−xx3→−16&amp;nbsp;

Evaluate: \(\displaystyle\lim_{{{x}\to{0}}}{\frac{{{{\sin}^{{3}}{x}}-{x}^{{3}}\ {s}{g}{n}{\left({1}-{\left[{\frac{{{x}}}{{{{\sin}^{{-{1}}}{x}}}}}\right]}\right)}}}{{{x}{{\tan}^{{2}}{x}}{\sin{{\left(\pi{\cos{{x}}}\right)}}}}}}\) (Note that here sgn denotes

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