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$$\lim_{x\to 0}\frac{\sin^3x-x^3\operatorname{sgn}\left(1-\left[\frac{x}{\sin^{-1}x}\right]\right)}{x\tan^2 x\sin(\pi\cos x)}$$

(Note that here $\operatorname{sgn}$ denotes the Signum function and $[x]$ denotes the greatest integer less than or equal to $x$.)

The above simplifies to:

$$\lim_{x→0} \frac{\sin^{3}(x)-x^{3}}{x\tan^{2}(x)\sin(\pi(1-\cos x))}$$

then after further simplification i get:

$$=\lim_{x→0} \frac{6}{π}\frac{\sin(x)-x}{x^{3}}$$

$$=\lim_{x→0} \frac{6}{\pi}\frac{-1}{6}=\frac{-1}{π},$$

but in this method I have to calculate the limit of

$$\lim_{x→0} \frac{\sin(x)-x}{x^{3}}=\frac{-1}{6},$$

which is lengthy. Is there any other way to do this without calculating this limit.

kadoodle
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  • Set $t = \sqrt[3]{x}$, so that what you want to compute is $$ \lim_{t\to 0} \frac{\sin^3 t^3-t^3}{t} = \lim_{t\to 0} \frac{f(t)-f(0)}{t-0} $$for $f(t) = \sin^3 t^3-t^3$ (defined on $\mathbb{R}$). $f$ is differentiable at $0$, and you want to compute $f'(0)$. – Clement C. May 30 '18 at 20:48
  • I want to avoid this limit. Is there any other way to solve this using standard limits. – kadoodle May 30 '18 at 20:55
  • You can compute $f'$ for all $t$ relatively easily, from compositions and rules for derivatives. Otherwise, yes, it's "pretty easy" if you know Taylor expansions. – Clement C. May 30 '18 at 20:57
  • @kadoodle Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Jun 22 '18 at 21:08

1 Answers1

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We have that

$$\frac{\sin^{3}x-x^{3}}{x\tan^{2}x\sin(π(1-\cos x))} =\frac{x^3-\frac12x^5+o(x^5)-x^3}{x^3(x^2+o(x^2))\frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}} =\frac{-\frac12x^5+o(x^5)}{(x^5+o(x^5))\frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}} =\frac{-\frac12+o(1)}{(1+o(1))\frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}}\to \frac{-\frac12}{1\cdot\frac \pi 2\cdot 1}=-\frac 1 \pi$$

Without Taylor's expansion note that we have

$$\frac{\sin^{3}x-x^{3}}{x\tan^{2}x\sin(π(1-\cos x))} =\frac{\sin^{3}x-x^{3}}{x^5}\frac{1}{\frac{\tan^2x}{x^2} \frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}}$$

then since by standard limits the second part $\to \frac 2 \pi $ we can analize

$$\frac{\sin^{3}x-x^{3}}{x^5}=\frac{(\sin x-x)(\sin^2 x+\sin x \cdot x+x^2)}{x^5}=\frac{\sin x-x}{x^3}\frac{\sin^2 x+\sin x \cdot x+x^2}{x^2}$$

and since by standard limits $\frac{\sin^2 x+\sin x \cdot x+x^2}{x^2}\to 3$ we need to consider

$$\frac{\sin x-x}{x^3}\to -\frac 16$$

which can be solved as indicated here Are all limits solvable without L'Hôpital Rule or Series Expansion.

user
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  • Can we do this without using the Taylor series or any such expansion...or is it necessary?? – kadoodle May 30 '18 at 21:06
  • @kadoodle We can simplify to $$\frac{\sin^{3}(x)-x^{3}}{x\tan^{2}x\sin(π(1-\cos x))} =\frac{\sin^{3}(x)-x^{3}}{x^3\tan^2x \frac{π(1-\cos x)}{x^2}\frac{\sin(π(1-\cos x))}{π(1-\cos x)}}$$ and then solve that $$\frac{\sin^{3}(x)-x^{3}}{x^3\tan^2x }$$ I add something on that. – user May 30 '18 at 21:14
  • But then I'll have to calculate the limit I don't want to... :-( – kadoodle May 30 '18 at 21:22
  • @kadoodle We can't avoid that! – user May 31 '18 at 05:58
  • okay............ – kadoodle May 31 '18 at 15:18