# Why is this limit said to equal some value rather than approach that value?

I have rewritten this entire question, since what I've learned since asking it requires me to restate it. I want to get rid of the obfuscating revisions.

Let's say that f is a continuous function.

$f(x)$ approaches L as x approaches a. So $\lim\limits_{x \to a}f(x) = L$

When it's said that the gradient of a tangent line to a curve at some particular point has some particular value, this is the same as saying f(a)=L. But without explicitly evaluating at that point, you can't say as much. All you can say is what happens as you approach that value. In other words, you can only say what value $f(x)$ approaches as $x$ approaches $a$, you can't say what $f(a)$ is.

Is that true?

Some textbooks will just say that the value of $f(a)=L$. Sal Khan's explanation does this. He says, about the function as it approaches the limit, "this is the gradient of the tangent." I'm saying that it should be said that, "the derivative approaches the gradient of the tangent to the curve." I think "approaches" and "is" are very different.

If there is a way to prove that the $f(a)=L$ then I'd like to see it. I don't know how to do this yet.

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Since "=" is "equals", it says the limit is L. Exactly L. (By the way, one cannot assume that $f(x)$ is continuous, because $f(x)$ is a number and a number is neither continuous nor discontinuous. One could assume that the function $f$ is continuous.) – Did Jan 4 '12 at 18:23
@Didier: the part of your remark about "$f(x)$" is technically correct but on the pedantic side. At least in the United States, speaking of "the function $f(x)$" is widespread practice among textbooks and instructors as well as students. We shouldn't say it, but we often do, and it doesn't (necessarily!) mean that we don't know the difference between a number and a function. – Pete L. Clark Jan 4 '12 at 18:25
@Didier, I actually appreciate your pedantry, I hadn't thought of it like that before. Now that you say it, it seems obvious. I will edit to clarify. My questions still stands, however. – Korgan Rivera Jan 4 '12 at 18:32
The phrase "$\lim_{x\to a} f(x)$ approaches $L$" conveys absolutely the wrong impression. A mathematician would never use it. One can say, correctly albeit informally, that $f(x)$ approaches $L$ as $x$ approaches $a$. The limit is (in this case), just a plain number, not at all a forever dynamic or moving process. Think of "She has arrived in Winnipeg." Winnipeg is a fixed place, the fact that she travelled to get there does not change that. – André Nicolas Jan 4 '12 at 19:12
Is the curve you're considering the graph $y=f(x)$? If so then the gradient of the tangent to the curve at $a$ would be $f'(a)$, not $f(a)$. – Adina G Jan 4 '12 at 21:23

The confusion seems to be in your line

$\lim\limits_{x\to a}{f(x)}$ approaches $L$.

It is $f(x)$ that approaches $L$.

That is, the function $f(x)$ approaches some value as $x$ tends to $a$. That value is called the limit, and is denoted $\lim\limits_{x\to a}{f(x)}$. Saying $\lim\limits_{x\to a}{f(x)}=L$ is saying these two values are equal.

The derivative $\frac{dy}{dx}$ is the gradient of the tangent line, which is the limit of gradients of successively smaller secants. Again, a limit is a value, not something that approaches a value.

Edit: let me see whether I can help with your rewritten question.

Suppose $f$ is a continuous function. Then $\lim\limits_{x\to a}f(x) = f(a)$ (this is by definition of continuity -- we haven't taken derivatives yet).

Now suppose $f$ is differentiable at $a$, and suppose $f'(a) = L$ -- that is, $$\lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h} = L$$.

The function $$g(h) = \frac{f(a+h)-f(a)}{h}$$, which is defined for $h$ non-zero, represents the gradients of secants. The function $g$ approaches $L$ as $h$ tends to $0$ -- this is what it means for $f$ to be differentiable at $a$. So $g$ approaches the gradient of the tangent to the curve at $a$. But $g$ is not the derivative. The derivative is $\lim\limits_{h\to 0}g(h)$, which is $L$.

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So what I should have said was $f(x)$ approaches $L$ as $x$ approaches $a$. So $\lim\limits_{x\to a}{f(x)}=L$ Some texts then go on to say, however, that $f(a)=L$. Isn't that the assumption being made when saying that the gradient of a tangent line where $x=a$ is $L$? – Korgan Rivera Jan 4 '12 at 20:06
@Korgan, it is not correct to say that $f(a)=L$ because $L=\lim_{x\to a}f(x)$. For any particular $f$ and $a$ it may of course happen to be true anyway. Or may not. – Henning Makholm Jan 4 '12 at 20:12
I see my mistake now. Thanks. – Korgan Rivera Jan 5 '12 at 1:54

This is the mere definition.

Letting $x$ approach $a$ then your function $f$ approaches some value $L$. You want to give this value a name. And you call it the limit of $f$ as $x$ goes to $a$, written $\lim_{x \to a} f(x)$.

Of course now $f$ approaches $\lim_{x \to a} f(x)$.

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When $x_n$ approaches some value $x$ as $n$ goes to infinity, then we say that the limit of $x_n$ is that value $x$. The limit is a number. A sequence approaches its limit (if it has one).

So you can't say "I understand that $\lim\limits_{x \to a}f(x)$ approaches $L$", since the limi $\lim\limits_{x \to a}[f(x)]$ is a number*, and the thing inside $\lim[\cdot]$ - in that case $f(x)$ - approaches $L$.

Coming to your question about existence of a limit.

(example 1) If the function $g:X\to Y$ is continuous and $a \in X$, then the limit $\lim_{x\to a}g(x)$ exists and is equal to $g(a)$. So: "when $x$ approaches $a$ then $g(x)$ approaches $g(a)$".

However it is possible that a function $h$ is not continuous at $a$, but still $\lim_{x\to a}h(x)$ exists.

(example 2) You can think about $h:X\setminus \{a\} \to Y$ such that $h(x)=g(x)$ from the previous example. If $a$ is not in the domain, we can't say about continuity there, but the limit obviously still exists.

(example 3) You can imagine $j(x)=\frac{x^2}{x^2+1}$ which has limits at infinity, so when $x$ approaches infinity, then j(x) approaches the limit, which is equal to $1$. There is no continuity here unless you know how to add infinities to your domain and define continuity for the enlarged domain.

(example 4) You can define a function $k$ equal to zero everywhere except for 0 where it is equal to one. It is artificial, but since in the typical definition of $\lim_{x\to a}$, one assumes that $x$ approaching $a$ is never precisely equal to $a$, then our function has limit at $0$ $\lim_{x\to 0}k(x)=\lim_{x\to 0}0=0$ but it (I mean the limit) is not equal to $k(0)=1$.

There is no such thing as "value at the limit". We just say that, since it is very handy. It is handy, because if the function is not defined in the limit point $a$, we may want to define it there, so then we add/define "value at the limit" to our function using its limit at this point.

(example 5) To produce $g$ (example 1) out of $h$ (example 2) we put $g(x) = h(x)$ for $x$ belonging to the domain of $h$ and $g(a) = \lim_{x\to a} h(x)$

*in the same way as $f(x)$ is a value, not a function

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## protected by Qiaochu YuanJan 5 '12 at 0:10

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