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This is a question from a calculus sample test, and I can't figure out how to prove it. Can I get some help from you guys?

Definition of continuity that we've learned is $$\lim_{x\to a} f(x) = f(a).$$ If that holds, then $f$ is continuous at $a$.

The definition that we learned of a limit is:

For every $\epsilon > 0$, there exists $\delta > 0$ such that $|x - a| < \delta$ implies $|f(x) - L| < \epsilon$.

$\epsilon$ and $\delta$, as far as I can tell, are just variables, $a$ is what $x$ is approaching, and $L$ is the limit.

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Is your definition of continuity at $a$ that $\lim_{x \to a} f(x) = f(a)$? You should mention this, as well as your question and what you've tried, in the body of the post for better results. I do think you'll get some excellent answers any moment now, though. –  Dylan Moreland Oct 1 '11 at 18:10
Yes, this sort of theorem depends on your definitions,and usually follows straight from them. In particular, it depends on the definition of "limit" and the definition of "continuous at a.". Different books use different definitions, so it is hard to give a proof without these definitions. –  Thomas Andrews Oct 1 '11 at 18:15
OK. What you need to show is $\lim_{x\to a}f(x)=f(a)$ if and only if $\lim_{h\to 0}f(a+h)=f(a)$. Do you know what does "if and only if" mean? Furthermore, do you know the definitions of these two limits? –  Jack Oct 1 '11 at 18:16
I know what if and only if means in English, and I think it means something very similar in math. I'm editing in the definitions we learned now. –  Lman Oct 1 '11 at 18:19
@Lman: Even "if" alone means something subtly different in mathematics than it does in English. For example, "Paris is the capital of France if $x^2<0$" is close to nonsense in ordinary English, but in math it is meaningful and true. –  Henning Makholm Oct 1 '11 at 18:43
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3 Answers

Have you tried the obvious guess, i.e. writing $x$ as $a+(x-a) = x$, and then noticing that $x \to a$ if and only if $h = x-a \to 0$?

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OK. An obvious step you should take is plugging the definition into you question:

$$\lim_{x\to a}f(x)=f(a)\qquad \text{if and only if} \qquad \lim_{h\to 0}f(a+h)=f(a)$$

As Gowers recently said, I think this is a fake difficulty for you. In order to answer your question, I would like to ask you more basic questions:

Do you know what does "if and only if" mean? Do you know the definitions of these two limits?

Now you can go on by yourself.

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Okay. Thanks for this, I think I'm making progress. If and only if means that one of them can't possibly be correct without the other one being correct as well, right? And if the definition on the left is correct, you can just plug in what the limit is approaching for the variable that is is approaching. The x becomes a. So assuming that's true, you can plug 0 in for h on the left side of the if and only if, and f(a+0) = f(a) –  Lman Oct 1 '11 at 18:43
@Lman: While what you say is correct as far as "if and only if" goes, perhaps it is better to think of it as giving you two implications: if the left hand side is true, then the right hand side has to be true as well; and (separately), if the right hand side is true, then the left hand side must be true as well. –  Arturo Magidin Oct 1 '11 at 21:02
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Here is my proof. Basically as others have said whenever you see "if and only if" you want to the statements imply each other.

So suppose $f$ is continuous at $a$, then $\forall \epsilon > 0$, $\exists \delta > 0$ such that $|x - a| < \delta \implies |f(x) - f(a)| < \epsilon$. In particular we want to show $\displaystyle\lim_{h\to0} f(a+h ) =f(a)$, so let us consider $|h - 0| < \delta$. Then we have $|h - 0| = |h| = |h + a - a| <\delta \implies |f(h + a) - f(a)| < \epsilon$ (here we sneakily let $x = h + a$) as desired.

On the other hand, let us assume we have $\displaystyle\lim_{h\to0} f(a+h ) =f(a)$ instead. Then $\forall \epsilon > 0, \exists \delta >0$ such that $|h| < \delta \implies|f(a+h) - f(a)| < \epsilon$. Consider $|x - a| < \delta \implies |f(x) - f(a)| = |f(a + (x - a)) - f(a)| < \epsilon$. Thus we have shown that the two are indeed equivalent (again we let $h = x-a$)

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