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$f(x)=\begin{cases}2x\,\,\,\text{if $x\geq 0$}\\-3x\,\,\,\text{if $x<0$}\end{cases}$ Show that $f(x)$ is continuous.


To show $f(x)$ is continuous, since I haven't covered $\epsilon-\delta$ definition, I would try to break it into three cases and use sequences: 1, when $x_0>0$;2, when $x_0<0$ and 3, when $x_0=0$ where $x_0\in\mathbb{R}$

Let $\{x_n\}$ be a sequence of $\mathbb{R}$

Case 1: Since $\{x_n\}$ is a sequence of $\mathbb{R}$, then its limit is in $\mathbb{R}$. So we could have $\lim\limits_{x\rightarrow\infty} f(x_n)=2x_0=f(x_0)$

Case 2: Using the similar way in case 1, we could get $\lim f(x_n)=-3x_0=f(x_0)$.

Case 3: Let $\epsilon >0$, since the limit of $\{x_n\}$ is in $\mathbb{R}$, then we could have $$|f(x_n)-f(x_0)|<\epsilon\tag1$$


For the solution above, I don't think I provide enough information to show $f(x)$ is continuous. And at $(1)$, I know that I need to use epsilon to show the function is continuous at $x_0$ but I don't know how use it. Can anyone give me a hit or suggestion to finish this problem? Thanks

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  • $\begingroup$ Do you need help to finish your way or any other way is okay for you? $\endgroup$ – Megadeth Oct 20 '15 at 2:55
  • $\begingroup$ @GudsonChou I need help to finish the way I am using. I haven't covered any new idea beside using sequences. $\endgroup$ – Simple Oct 20 '15 at 2:57
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Hint:

$f$ is continuous at $x=0$ if $\lim_{x\to 0}f(x)=f(0)$, and $\lim_{x\to 0}f(x)$ exists if and only if $\lim_{x\to 0^+}f(x)=\lim_{x\to 0^-} f(x)$.

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  • $\begingroup$ Thanks for the hint. I wander how could use $\epsilon$ to show the function is continuous at $0$. $\endgroup$ – Simple Oct 20 '15 at 3:13
  • $\begingroup$ Find a $\delta$ that works for the function $2x$ and a $\delta$ that works for the function $-3x$. The smaller of the two works for both. This is essentially an implementation of the idea I gave in my answer. $\endgroup$ – Alex S Oct 20 '15 at 3:15
  • $\begingroup$ I haven't covered $\epsilon-\delta$ definition $\endgroup$ – Simple Oct 20 '15 at 3:16
  • $\begingroup$ Then consider an arbitrary negative sequence converging to $0$ and a positive sequence converging to $0$. Use $\varepsilon$ to show that the function applied to both of these sequences converges to $0$. This is another way to apply the above idea. $\endgroup$ – Alex S Oct 20 '15 at 3:19
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    $\begingroup$ The idea is right, but it is wrong to say that $3|x_n|=0$. It is true that for any $\varepsilon>0$ and $n$ sufficiently large, $3|x_n|<\varepsilon$. Also, you should write $|f(x_n)-f(x_0|\leq 3|x_n|$, since they may not be equal for positive $x_n$. $\endgroup$ – Alex S Oct 20 '15 at 3:46

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