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We consider $f_n(x)=x^n$ on $[0,1]$. Each function $f_n(x)$ is continuous, but the limit function $f(x)$ is not continuous: $$ f(x)=\left\{ \begin{array}{ll} 0, 0\leq x<1\\ 1, x=1\\ \end{array} \right. $$

Question 1: How can we prove that: $$ \lim_{n\rightarrow \infty}(\lim_{x\rightarrow 1}f_n(x))=1 $$ $$ \lim_{x\rightarrow 1}(\lim_{n\rightarrow \infty}f_n(x))=0 $$

Question 2: Is uniform convergence of $f_n(x)\rightarrow f(x)$ a sufficient condition to get the same result if we interchange the limits?

1 Answers 1

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Question 1:

$$\lim_{x\to 1} f_n(x) = \lim_{x\to 1} x^n = 1^n = 1$$

because, for every $n$, the function $x\mapsto x^n$ is continuous.

Therefore,

$$\lim_{n\to\infty} \lim_{x\to 1} f_n(x) = \lim_{n\to\infty} 1 = 1$$

On the other hand, if $x<1$ (we can assume this when calculating $\lim_{x\to 1}$), then

$$\lim_{n\to\infty} f_n(x) = \lim_{n\to\infty}x^n = 0$$

So $$\lim_{x\to 1}\lim_{n\to\infty} f_n(x) =\lim_{x\to 1} 0 = 0$$

Question 2: Yes.

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    Thank you for the reply! What I didn't get by myself is the assumption $x<1$, now it's clear. Regarding question #2: Is there any theorem that addresses such property? – 2017-01-05
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    @Konstantin Are the functions $f_n$ continuous? – 2017-01-05
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    Can you give some hints for both continuous and discontinuous cases? – 2017-01-05
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    @Konstantin Well, if $f_n$ are continuous, then $f$ (the limit function) is continuous by a theorem, and everything just falls out because $\lim_{x\to x_0} f_n(x)=f_n(x)$. Not sure about when $f_n$ are discontinuous. Will think about it. – 2017-01-05
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    So we apply uniform limit theorem and `f` is continuous. But why everything "just falls out"? – 2017-01-05
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    @Konstantin Yeah, since $f(x)$ by definition is $\lim_{n\to\infty} f_n(x)$, you get that $$f(x_0) = \lim_{n\to\infty} f_n(x_0) = \lim_{n\to\infty} \lim_{x\to x_0} f_n(x)$$ because $f_n$ are all continuous. But you also know that $f$ is continuous, so you have $$f(x_0)=\lim_{x\to x_0} f(x) = \lim_{x\to x_0}\lim_{n\to\infty} f_n(x)$$ – 2017-01-05