0
$\begingroup$

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.

  • 0
    @Lman: Even "i$f$" alone means something subtly different in mathematics than it does in English. For e$x$ample, "Paris is the capital of France **if** $$x$^2<0$" is close to nonsense in ordinar$y$ English, but in math it is meaning$f$ul and true.2011-10-01

4 Answers 4

4

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 $h = x-a \to 0$?

  • 0
    Your question doesn't quite make sense either. I don't even see a question in the OP's post anymore, it must have been edited. Of course$x$-> a if and only if x-a -> 0, that doesn't need proof.2018-01-11
2

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.

  • 1
    @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.2011-10-01
0

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 $h = x-a$)