I found this procedure:
for $x \in (0,1) \ \ \lim_{n\to\infty} nx(1-x^4)^n = x \cdot lim_{n\to\infty} n(1-x^4)^n = 0$
But why is it evaluated to $0$? Wouldn't it lead to $0 \cdot \infty$ (as $n$ multiplies $(1-x^4)^n$ , which is equal to $0$ in the interval $(0,1)$)?
2026-04-04 14:51:50.1775314310
How do you find that $\lim_{n\to\infty} nx(1-x^4)^n$ is equal to $0$ for $x \in (0, 1)$?
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It is not "evaluated at zero". Here $x $ is a fixed number between $0$ and $1$. So it can be "pulled out" from the limit. Also, as $0 <x <1$, we have $0 <1-x^4<1$. Then $\log (1-x^4)<0$, and $$ n (1-x^4)^n=ne^{n\log (1-x^4)}\to0. $$ Multiplied by the constant $x $, it still goes to,zero.