How to interpret limit notation

Question

$\lim\limits_{x \to a} f(x)= L$ is by most; intuitively thought of "as $x$ gets close to $a$, $f(x)$ gets close to $L$", however my lecturer said this is not correct. She told me to go away and somehow find out why, by formal definition, the intuition "$f(x)$ is close to $L$, for all $x$ sufficiently close to $a$" is correct, not the former.

I went on to find examples; Simply consider; $f(x) = x/|x|$ when $x$ tends to some number. and to recall an emphasise; “As $x$ gets close to $a$, $f(x)$ gets close to $L$”

The emphasise on gets is important as it suggests some change towards $L$, however when investigating, as $x$ tends to some number (like $0$), $f(x) = L$, no matter where on the domain you fly. There ceases to be a case in this function where $f(x)$ moves/gets close to $L$ anywhere.

“$f(x)$ is close to $L$, for all $x$ sufficiently close to $a$” includes the idea of ‘there exists some interval’ where $f(x)$ is close to $L$.

Is that a sufficient answer to the question? I can't find anything online.

Answer 1

It is wise to tread carefully when semantically wording some intuition of a precise definition.

However; in this particular case there exists at least one function that smashes the first and is consistent with the latter. That is taking the limit as $x$ tends to $0$ of;

$f(x) = xsin(1/x)$.

The statement "as x gets close to a, f(x) gets close to L" simply does not hold because no matter how close to $a$ you get, your function continues to oscillate towards and away from L. However by definition the limit does exist.

Answer 2

I think that it is easier to understand the negation.

$f (x) $ could never be closer to $L $ $(|f (x)-L|>\epsilon) $

even if $x $ is closer to $a \;(|x-a|<\delta) $

Answer 3

The epsilon-delta definition is pretty straight-forward:

$$\lim _{{x\to c}}f(x)=L\iff (\forall \varepsilon >0)(\exists \ \delta >0)(\forall x\in D)(0<|x-c|<\delta \ \Rightarrow \ |f(x)-L|<\varepsilon )$$

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What does this mean? Well, we break it down, part by part:

1. $(\forall\varepsilon>0)\dots(\dots|f(x)-L|<\varepsilon)$

This means that $f(x)$ can get arbitrarily close to $L$.

2. $(\exists \ \delta >0)(\forall x\in D)(0<|x-c|<\delta\dots)$

This means that the previous statement is true for every $x$ in the domain that is a certain distance from $c$, the value $x$ is approaching.

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This is different from your definition in that it requires $f(x)$ to be close to $L$ with some maximum error $\varepsilon$ for all values $x$ close to $c$. $x$ does not merely approach $c$, but instead, we must have

$$|f(x)-L|<\varepsilon$$

for every $x$ close to $c$. The next requirement would then be that the distance between $f(x)$ and $L$ can keep getting smaller and smaller, and that it would still hold for every $x$ values a certain distance from $c$.

There is no such "$x$ approaches $c$" here.

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So yes, $f(x)$ is close to $L$ for all $x$ sufficiently close to $a$ is the accurate statement.

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In the example $x/|x|$, notice that no matter what $x$ value you choose, either the result will be $1$ or $-1$. Let's imagine taking the limit to $10$.

$$\lim_{x\to10}\frac x{|x|}\stackrel?=1$$

We then make a table of values:

$$\begin{array}{c|c}x&f(x)\\\hline9&1\\9.9&1\\9.99&1\\\vdots&\vdots\\10.01&1\\10.1&1\\11&1\end{array}$$

Notice that $f(x)$ does not approach anything. It is simply constant. I suppose you could then try to fix your statement with "as $x$ approaches $10$, $f(x)$ is close to $1$ within some amount of error that tends to zero."

But that misses the intuition you can get with the epsilon-delta definition:

Notice that $f(x)-L=0\forall x>0$. It thus follows that $|f(x)-L|=0<\varepsilon$, which holds when $\delta=10$. $\delta$ needn't get smaller. It simply needs to be small enough.