I need to compute the limit or state it doesn't exist for: $\lim_{x\rightarrow 0} x^{\frac{1}{3}} (-1)^{\big[\big[\frac{1}{x}\big]\big]}$

I have already computed that: $\lim_{x\rightarrow 0} (-1)^{\big[\big[\frac{1}{x}\big]\big]}$ does not exist. Since if we let $x_n = \displaystyle\frac{1}{2n+1}$ and $y_n=\displaystyle\frac{1}{2n}$ we have that $\lim_{n\rightarrow\infty} x_n = \lim_{n\rightarrow\infty} y_n = 0$ but $\lim_{n\rightarrow\infty} (-1)^{\big[\big[\frac{1}{x_n}\big]\big]} \neq \lim_{n\rightarrow\infty} (-1)^{\big[\big[\frac{1}{y_n}\big]\big]}$.

How would I now compute the limit above?

  • $\begingroup$ Take its absolute value and use the very definition of limit $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 19 '17 at 4:36
  • $\begingroup$ As the question is worded, you can just state that it doesn't exist. It doesn't seem to ask that the statement be true, just that you take one of two actions. $\endgroup$ – Ross Millikan Dec 19 '17 at 4:45

Noting that as $x \to 0, (-1)^{\lfloor\frac{1}{x}\rfloor}$ is either $-1 $ or $1$, ie both finite values. But as $x\to 0, x^{1/3} \to 0$.

So the overall limit must be $0$ (product of finite value and a number tending to zero).


Note that

$$0\leq \left| x^{\frac{1}{3}} (-1)^{\big[\big[\frac{1}{x}\big]\big]}\right|\leq x^{\frac{1}{3}}$$

thus by sqeeze theorem

$$x^{\frac{1}{3}} (-1)^{\big[\big[\frac{1}{x}\big]\big]}\to 0$$


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