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$\{f_n\}$ is a sequence of continuous functions such that $f_n \to f$ uniformly on $\mathbb{R}$. Suppose that $x_n \to x_0$, prove that $\displaystyle \lim_{n \to \infty} f_n (x_n) = f(x_0)$.

I think I must be missing something here, but I don't see how this is a rigorous question beyond intuition. If a sequence of functions converges to some form, and the variable placed in the function converges to a value, it only makes sense that the function value should also converge accordingly.

Could someone give hints on how to approach this rigorously?

Edit: could someone clarify the conclusion in the last sentence of the answer comes to be?

kiwifruit
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1 Answers1

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We know that we can make the term $|f_n(x_n) - f(x_n)|$ small since $\{f_n\}$ converges uniformly. We know that we can make $|f(x_n)-f(x_0)|$ small since $f$ is continuous, and $x_n \to x_0$. Using this information, we can make $|f_n(x_n)-f(x_0)|$ small, which is what we want to do to show that $\lim f_n(x_n) = f(x_0)$.

Edit: To be more specific, let $\varepsilon >0$. Choose $N_1$ such that $|f_n(x_n)-f(x_n)|<\varepsilon/2$ whenever $n \geq N$. (Why can we do this?) Choose $\delta>0$ such that, whenever $|x - x_0|<\delta$, we have $|f(x)-f(x_0)|<\epsilon/2$. (Why can we do this?) Choose $N_2$ such that $|x_n - x_0| < \delta$ for $n \geq N_2$. Finally, let $N = N_1 + N_2$. I claim that for $n \geq N$, we have $|f_n(x_n)-f(x_0)| < \varepsilon$. Can you see why?

Nick
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