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So, I'm stuck with this proof for a homework assignment, and don't know where to start - my first instinct is to go with a proof by squeeze theorem. Would that work here though? I'm just not sure right now. Here's the question:

Suppose $g$ and $h$ are both defined as being in the neighborhood of $c$ and

$$\lim_{x\to c}g(x)=L=\lim_{x\to c}h(x)\;.$$

Define $f$ by

$$f(x) =\begin{cases} h(x),&\text{if }x\in\Bbb Q\\ g(x),&\text{if }x\in\Bbb R - \Bbb Q\;. \end{cases}$$

Show that $\lim\limits_{x\to c} f(x) = L$.

Community
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Barney
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1 Answers1

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HINT: Just use the $\epsilon$-$\delta$ definition of limit. Given $\epsilon>0$, you know that there are $\delta_g>0$ and $\delta_h>0$ such that $|g(x)-L|<\epsilon$ whenever $0<|x-c|<\delta_g$ and $|h(x)-L|<\epsilon$ whenever $0<|x-c|<\delta_h$. Let $\delta=\min\{\delta_g,\delta_h\}$; what can you say about $|f(x)-L|$ when $0<|x-c|<\delta$?

Brian M. Scott
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  • So, since f(x) is comprised of g(x) and h(x), then it too would be less than epsilon, given that the delta is the minimum value between g(x)'s and h(x)'s deltas? – Barney Nov 22 '13 at 17:41
  • @Barney: Exactly. If $0<|x-c|<\delta$, then $0<|x-c|<\delta_g$ and $0<|x-c|<\delta_h$, so no matter whether $x$ is rational or irrational, you have $|f(x)-L|<\epsilon$. – Brian M. Scott Nov 22 '13 at 17:51