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Let the functions $f_n (x) = x^{n+1} - x^n$. Prove that the limit $\lim \limits_{x \to 1} f_n (x) = 0$, and the limit is uniform on $n$, i.e., given any $\varepsilon > 0$, there exist a general $\delta > 0$ that depends only on $\varepsilon$, such that $$ |x - 1| < \delta \ \Rightarrow |f_n(x) - 1| < \varepsilon $$ for every $n$.

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I don't believe this statement when $x > 1$. $f_n(x) = x^n(x-1)$ and for any given $x$ you can take $n$ large enough that $x^n$ is arbitrarily large. Perhaps I am wrong though. –  badatmath Nov 17 '11 at 18:33
    
That´s what I think D: –  Susuk Nov 17 '11 at 18:35
    
Maybe it was only supposed to be proven for $x < 1$? –  badatmath Nov 17 '11 at 18:38
    
Or the book it´s wrong, but at least in this case , is true . And yes, you are right, clearly the other side not –  Susuk Nov 17 '11 at 18:40

1 Answer 1

The result is false as stated.

Suppose the statement is true, and let $x = 1 + \delta$. There exists an $n$ such that $(1 + \delta)^n$ is greater than $\epsilon/\delta$, and therefore such that $(1 + \delta)^n((1 + \delta) - 1) > \epsilon$.

This claim is true because suppose the sequence $((1 + \delta)^n)_{n = 0}^{\infty}$ was bounded by a bound $M$. Then by the completeness axiom it has a least upper bound $S$. There must exist a term of the sequence that is in the interval $(\frac{S}{1 + \delta}, S)$ (or else $\frac{S}{1 + \delta}$ would be a smaller upper bound). Let $(1 + \delta)^n$ be that term. Then $(1 + \delta)^{n+2} > S$, and the sequence cannot have an upper bound.

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