I am trying to find the limit as $x \rightarrow \infty$ (x is an integer not a continuous variable) of the following

$(1 - x^\alpha)^x $

knowing that $0 < \alpha < 1$. Mathematica tells me its limit is 0 but cannot find a way by hand. my attempt involves writing the formula as a binomial expansion but I get the limit is 1 ...

Any help would be appreciated, thank you

  • $\begingroup$ Please show how you got one as limit. $\endgroup$ Nov 8, 2019 at 15:03
  • $\begingroup$ clearly the limit does not exist as it will tend to negative infinity for odd $x$ and to positive infinity if $x$ is even $\endgroup$
    – Vasili
    Nov 8, 2019 at 15:13
  • $\begingroup$ @ViktorGlombik: Thanks, I missed that... we ought to outlaw the use of $x$ as a discrete variable :) $\endgroup$
    – Clayton
    Nov 8, 2019 at 15:37

1 Answer 1


If $x > 1$ then $x^\alpha > 1$ and $1-x^\alpha < 0$ so that $(1-x^\alpha)^x$ changes sign as $x$ is even or odd.

Moreover $|1-x^\alpha|^x \ge |1-x^\alpha| = x^\alpha - 1$ for all $x \ge 1$. Since $x^\alpha \to \infty$ you conclude your sequence diverges rather badly.

  • $\begingroup$ @Clayton Of course, of course. Please forget my complaint. $\endgroup$
    – Allawonder
    Nov 8, 2019 at 16:14

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