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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.2011-11-17
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    That´s what I think D:2011-11-17
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    Maybe it was only supposed to be proven for $x < 1$?2011-11-17
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    Or the book it´s wrong, but at least in this case , is true . And yes, you are right, clearly the other side not2011-11-17

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