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I'm asked to show that the sequence of functions $f_n(x) = n^2x^n$ defined on the closed interval $[0,1]$ does not converge pointwise to any function as $n \to \infty$.

For $0 \le x \lt 1$ I think I have convergence to zero, but for $x = 1$ the sequence of functions is clearly divergent. I conclude, from the question, that this divergence is sufficient to conclude that the sequence does not converge to any function. But I have a feeling that the function $f(x) = 0$ for $0 \le x \lt 1$ is a valid function, doesn't the sequence converge to it?

Also, the sequence $$g_n(x) = \frac{1}{\frac{n+1}{n}-x}$$ is clearly divergent at $x = 1$, but also converges to $g(x) = \frac{1}{1-x}$.

What am I missing?

user164587
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    you are right: the limit function isn't "defined" at $x=1$, so the sequence of functions doesn't converge to any function on the whole interval: even not converging at a point would cause a sequence of functions not to converge. – voldemort Sep 07 '14 at 02:58
  • @voldemort How does this differ to my example with $g_n(x)$ converging to $g(x)$? Do you disagree that the $g_n$ converge to $g$? Each $g_n$ is defined on $[0,1]$ but the limit function is not. Neither is $g$. – user164587 Sep 07 '14 at 03:31
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    $g_n(x)$ doesn't converge to $g(x)$ on $[0,1]$ – voldemort Sep 07 '14 at 03:33
  • Great, thanks! That's exactly what I supposed the answer was. – user164587 Sep 07 '14 at 03:48

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