# The limit of $(1 - x^\alpha)^x$ as $x \rightarrow \infty$

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

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

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.