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If $f$ is a function then $$\lim_{x\to a}f(x) = L$$

implies that $f(x)$ gets closer to $L$ as $x$ gets closer to $a$.

Is this right?

TheWanderer
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stadigfora
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  • Well, I would rather use the verb 'means' in place of 'implies', but yes, at least in a small neighborhood of $a$. And this intuitive meaning has a precise definition. – Berci Mar 09 '18 at 21:23
  • In broad strokes that is the idea, but it is not a rigorous definition. It is good enough to get you through a first year calculus class, but If you go on to take real analysis, you will need to more thoroughly understand a more precise definition. – Doug M Mar 09 '18 at 21:47
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    You can improve your idea of limit by saying that values of $f$ can be ensured to lie as near to $L$ as we please for all values of $x$ which lie sufficiently close to $a$. – Paramanand Singh Mar 09 '18 at 22:01
  • @ParamanandSingh You could add an answer on this! – user Mar 09 '18 at 22:05
  • @gimusi : I have written about this in past. See https://math.stackexchange.com/a/1324644/72031 – Paramanand Singh Mar 10 '18 at 16:55
  • @gimusi : and another one here https://math.stackexchange.com/a/1733170/72031 – Paramanand Singh Mar 10 '18 at 17:07
  • @ParamanandSingh Thanks a lot! – user Mar 10 '18 at 17:09
  • @stadigfora Please remember that you can choose an aswer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 12 '18 at 23:44

3 Answers3

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That is what it means, yes.

And even better: $f(x)$ can come arbitrarily close to $L$ as $x$ gets closer to $a$. This is a subtle, but important difference: $\cos(x)$ gets closer and closer to $2$ as $x$ gets closer to $0$, but you can never actually get really close to $2$. So $\lim_{x\to0}\cos(x)$ isn't $2$.

Also important: as $x$ gets closer to $a$, $f(x)$ not only gets close to $L$, but it gets close to $L$ for all values of $x$ close to $a$. For instance, $\sin(1/x)$ gets close to $0$ as $x$ gets close to $0$, but it is also far away for some $x$ close to $0$ as well. So $\lim_{x\to0}\sin(1/x)$ isn't $0$.

Arthur
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Yes it is the naive idea for the definition of limit for which get closer means "get eventually and arbitrarly closer" with respect to any prescribed bound but not necessarly equal to the limit and not necessarly monotonically, that is formally

$$\left(\lim_{x\rightarrow a} f(x) = L\right) \iff \Big(\forall \varepsilon >0\, \exists \delta > 0: \big(0<\vert x-a\vert <\delta \implies \vert f(x)-L\vert <\varepsilon\big)\Big)$$

user
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  • Thus the official definition od limit is wrong? That's good to know! We never end to learn! – user Mar 09 '18 at 21:42
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    I initially downvoted because to me "[eventually] $f(x)$ gets close to $L$ as $x$ gets close to $a$" does not imply "[eventually] $f(x)$ gets closer to $L$ as $x$ gets closer to $a$". But in retrospect, this language is ambiguous, so I reversed. – K B Dave Mar 09 '18 at 21:58
  • @KBDave Thanks for your comment, indeed it is not possible to guess what exactly the asker has in his mind and what he means exactly for "get closer", anyway the aswers given here will give to him/her a nice and complete overview that the naive "get closer" idea has behind more complex concepts which require a rigorous definition. – user Mar 09 '18 at 22:01
  • @ParamanandSingh Thanks! indeed it is often a matter of great confusion. – user Mar 10 '18 at 17:22
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No. Take$$f(x)=\begin{cases}x\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ if }x=0\end{cases}$$Then $\lim_{x\to0}f(x)=0$. However, no matter how close you are to $0$, there are points $x$ at which $f(x)=0$ and there are points $x$ at which $f(x)\neq0$.

José Carlos Santos
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  • ok there are points at which f(x)=0 but we get closer according to the definition of limit, I don't understand the initial "No" in the answer. Could you please explain? Thanks – user Mar 09 '18 at 21:30
  • @gimusi We don't get closer and closer. When $x=\frac1\pi$, $f(x)=0$. So, we couldn't possibly get closer to $0$ than that. But after that, as $x$ approaches $0$, we actually get away from $0$, and then we get again closer to $0$, reaching it again when $x=\frac1{2\pi}$. And then we get again away from $0$. And so on. – José Carlos Santos Mar 09 '18 at 21:35
  • Yes of course but, as we know, we can get arbitrarly and eventually closer as we want thus the initial "No" in your answer seems a little bit stringent to me since it depends upon what the asker mean with the term "get closer". – user Mar 09 '18 at 21:41
  • There is no $\delta>0$ such that, for all $\lvert x \rvert<\delta$, $\lvert f(x)\rvert$ decreases as $\lvert x\rvert$ decreases, even though for all $\delta>0$, $\max_{\lvert x\rvert \in[0,\delta]}\lvert f(x)\rvert$ decreases as $\delta$ decreases. – K B Dave Mar 09 '18 at 21:42
  • @gimusi Let us wait for the comments by the OP. – José Carlos Santos Mar 09 '18 at 21:43
  • @JoséCarlosSantos Ok, meanwhile note that the OP doesn't state "gets closer and closer" but simply "get closer" that is just the definition of limit, by the way the definition of limit that I know. – user Mar 09 '18 at 21:45