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For
$$f_n(x)= \left\{ \begin{array}{ll} n & x\geq n \\ 1 & x< n \\ \end{array} \right. $$ Graphically (is this usually how it's first approached), I can see that $f_n$ approaches to the function $f(x) = 1$ pointwise.
This is my proof (formal) for this limit, but I'm stuck on one part.
A set of functions $\{f_{n}\}_{n\in \mathbb{N}}$ converges to a function $f$ pointwise $$\iff \forall x \in I, \forall \epsilon > 0 \quad \exists N : n \geq N \implies |f_n (x) - f(x)| < \epsilon. $$ By inspection of the function $f_n$, we can see that picking $N = x$ implies that if $n \geq x$, then $x \leq n$. Now for the case that $x < n$, then $|f_n (x) - f(x)| = |1 - 1| = 0$ which is less than $\epsilon$ by definition.
However, I can't seem to figure out what if $x = n$? i.e. if $n = N$? This means I have to show that $|n - 1| < \epsilon$, which I'm not sure how to.

Natash1
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  • Pick $N = x+3$ instead, and your concerns should vanish. I understand the pedagogical need to pick a tight bound when teaching $\epsilon$-$\delta$-like stuff, so that the students can see what leads to what, but there is no reason to stick to that when you work on something for yourself. – Arthur Jun 02 '17 at 09:13
  • Haha thanks. Easy way to fix it (and I assume picking $N = x+a$ for any $a\geq 1$ is also valid).
    Is this the standard method to proving formal definitions of piece wise functions? It seems to be.     – Natash1 Jun 02 '17 at 09:18
  • Yes, any such $a$ will do fine, I picked $3$ at random. As for proving formal definitions on piecewise functions, I don't see how they're different from non-piecewise functions. It's just a function, after all. – Arthur Jun 02 '17 at 09:20
  • It feels like piecewise functions would be easier in general, since we "split" up the interval. So if we were to pick the interval that allows $f_n$ to attain the value of its limit function, then it'd be an easy pick. I might not have had much exposure to more problems yet – Natash1 Jun 02 '17 at 09:22
  • You say that just because the pieces the function is built form are extremely simple. Once they too start having exponentials, logarithms and rational functions like the non-piecewise examples you have probably seen, you will have to approach it like a normal function again. – Arthur Jun 02 '17 at 09:23

1 Answers1

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Let $x \in \mathbb R$. Then there is $ N \in \mathbb N$ such that $x<N$. For $n \ge N$ we then get

$x<n$ and therefore $f_n(x)=1$.

Result: to each $x$ there is $N$ such that $f_n(x)=1$ for all $ n \ge N$

Fred
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