What does limit actually mean?

Question

I have been in a deep confusion for about a month over the topic of limits! According to our book, the limit at $a$ is the value being approached by a function $f(x)$ as $x$ approaches $a$.

I have a doubt that in real number line we can never ever reach the closest value to $a$ because always a more closer value will exist.

Now when talking about our methods for calculating limit, what method comes in our mind when we have to calculate where the value approaches? Let's say we have to find what value $f(x)$ approaches when $x$ moves from $[0,a)$.

So we calculate the value of $f(x)$ at $x=0$, say $4$, and then at $x\to a$.

But the problem is that we don't know the value $x\to a$, so we say the value is $a-h$ where $h\to 0$ and calculate the value of $f(a-h)$, say $5-h$.

Now here is where my doubt starts! In the final step we put the value of $h=0$ say in that it is an infinitesimal quantity.

My doubt is that $h$ was tending to $0$ means that it was never equal to zero maybe it is infinitesimally small, not a stationary value, not an imaginable value, but we know for sure that it is not equal to $0$. Maybe it is the point closest to zero, but it is not equal to zero and when we use the result $5-h=5$ we are actually making an error which is tending to zero. Maybe the error is very small, but still, there is some error in that we cannot calculate it but we can see that there is this infinitesimal error present.

That means we don't get the exact limiting value or last value of $f(x)$.
$x$ belongs to $[0,a)$ but a value approximate to infinitesimal? Isn't it right! We get an approximated value?

Answer 1

So far as I can understand your question, you need to know that the Real numbers are set up in such a way that limits make sense (defined or constructed - depending on the approach - however they arrive they have a unique set of properties).

The Rational numbers do not behave well when we take limits - $\sqrt 2$ is the usual simple example - we can get as close as we like, but never equal. But $\sqrt 2$ exists as a Real number.

The Real numbers do not have lots of infinitesimals trying to get in the way. As always in mathematics, there are systems which do have infinitesimals, and which approach some of the issues in a different (non-standard) way. But at the heart of working with the Real numbers and with limits is using again and again those special properties of the Real numbers which were designed with limits in mind.

Answer 2

Maybe a simple discontinuous example will enlight you. Let $f(x) = 0$ if $x \leq 0$, $1$ if $x> 0$. I want to compute the right limit, that is $\lim_{x\to0,x>0} f(x)$.

In this case you are right, there is a difference between $f(0)=0$ and $f(\text{small quantity})=1$. The infinitesimal error induces a non infinitesimal difference in the values.

However this won't happen when the function is continuous. By definition, continuity means that f(x) equal to its limit $\lim_{dx\to0} f(x+dx)$ at every point. In this case, you can neglect the infinitesimal $dx$ and write $f(x) = \lim_{dx\to 0} f(x+dx)$.

The real issue here is that you would use continuity to prove something on the limits, while this limit is required to prove continuity itself. There exists a true mathematical definition of limits, but it seems out of the scope of your current grade.

The real definition of limit is "you can get as close as you like to the limit". Hence $\lim_{x \rightarrow x_0} = l$ actually means $\forall \epsilon, \exists \eta / \forall x, d(x,x_0) < \eta \Rightarrow d(l,f(x_0)) < \epsilon$, where d is some distance. The $\epsilon$ formalizes the "as close as you like' .I am afraid you can't do clean maths on limits without this definition

Answer 3

Well, in a sense, you're right. When they say that the limit of $f(x)$ at $x=a$ is $L$ it doesn't necessarily mean that $f(a)=L$. Actually, the nice idea behind limits is that you can talk about the limit of a function even if the function is not defined at that value. This is a very powerful idea that later enables us to talk about derivatives as you possibly know.

For example $\displaystyle \lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$ but the value of $\displaystyle \frac{\sin(x)}{x}$ is not defined at $x=0$. If you graph it on wolframalpha, you'll see that this means 'as we approach $x=0$ the value of $\displaystyle \frac{\sin(x)}{x}$ approaches 1'. We never claim that these two are equal! We just claim that the value of $f(x) = \displaystyle \frac{\sin(x)}{x}$ can become arbitrarily close to $1$ provided that we let $x$ be close enough to $0$ .

When we say that the limit of $f(x)$ at $x=a$ is $L$, we are claiming that we can make $f(x)$ arbitrarily close to $L$ provided that we take $x$ close enough to $a$. That's all.

Answer 4

I am not sure that I understand exactly what you are asking.

However, maybe the following might aide to your understanding. To truely understand limit you might want to look at the definition. That is: $$ \lim_{x\to a} f(x) = L $$ means that

for all $\epsilon >0$ there is some $\delta >0$ such that if $0<\lvert x-a\rvert < \delta$ then $\lvert f(x) - L\rvert < \epsilon$.

Now this might look intimidating, and in many ways it is. It will take some time to get your head wrapped around the definition, but if you are willing to do some work, I think that you can do it.

That said, while we have the definition to that tell us exactly what a limit is, there are ways to think about a limit. This "ways of thinking about a limit" can be helpful in getting an idea/picture of what limit means.

That $f(x)$ approaches $L$ as $x$ approaches $a$ means that we can make the values of $f(x)$ as close to $L$ as we would like by making $x$ as close enough to $a$.

You write that:

So we calculate value of $f(x)$ at $x=0,$ say $4$, and then at $x\to a$.

You also write

I have a doubt that in real number line we can never ever reach the closest value to a because always a more closer value will exist.

Note that we are not actually calculating the value of $f(x)$ at $x\to a$. That doesn't make sense. That $f(x)$ approaches $L$ as $x$ approaches $a$ doesn't mean that we can evaluate $f$ at some number and then get $L$. The values of $f(x)$ do not need to ever equal $L$. We are not evaluating $f$ at some "infinitesimal quantity".

Example: Consider the function $f(x) = x+7$. Then we obviously have that $$ \lim_{x\to 1} f(x) = 8. $$ In this example we even have that $f(1) =8$, but again this is irrelevant for finding the limit. It should be clear that when $x$ is very close to $a$ then $x + 7$ is very close to $8$.

Example: No consider this function: $$ g(x) = \begin{cases} 0 & \text{ if } x = 1 \\ x + 7& \text{ if } x \neq 1 \end{cases}. $$ Now we have that $g(1) = 0$, but we still have that $$ \lim_{x\to 1} g(x) = 8. $$ As $x$ approaches $1$ we don't have $x$ ever equal to $1$. So, wince $x$ is never equal to $1$ we can work with the expression $x+7$ for $g(x)$. So again $$ \lim_{x\to 1} g(x) = 8. $$ Example: One more example woulf be to consider the function $$ h(x) = 7. $$ This is the constant function that is equal to $7$ for all $x$. So what is the limit $$ \lim_{x\to 1} h(x)? $$ It is $7$ because we can make the values of $h(x)$ as close to $7$ as we want by making $x$ close enough to $1$. If you, for example, wanted the values of $h(x)$ to be with in $0.001$ (so here $\epsilon = 0.001$ in the definition) of $7$, you can take any tolerance for $x$ because $h$ is always equal to $7$.

Answer 5

I would like to respond to OP's comment to the effect that "Maybe the error is very small, but still, there is some error in that we cannot calculate it but we can see that there is this infinitesimal error present." In fact, discarding such an infinitesimal remainder is precisely what the limit operation does, in the following sense. In an extended number system, say $E$, including infinitesimals, there is a function from the finite elements of $E$, denoted $E_f$, to $\mathbb{R}$. This function is called the standard part function, and is denoted "st". Thus we get $$\text{st}:E_f \to \mathbb{R}.$$ For example, attempting to calculate the derivative of $y=x^2$ using the quotient $dy/dx$, one gets $2x+dx$. To pass from this to the expected formula $2x$, one applies the standard part: $\text{st}(2x+dx)=2x$. More generally, the limit of $f(x)$ as $x$ tends to $x_0$ can be defined as $\text{st}(f(x_0+dx))$, whenever the resulting value is independent of the nonzero infinitesimal $dx$ chosen.

The complications in the usual epsilon, delta definitions stem from a preference for working in $\mathbb{R}$ rather than a richer system containing infinitesimals, which is fine. However, the basic point about the limit operation is that it amounts to discarding the remaining term you referred to as "error".

So the answer to the question: "What does limit actually mean?" is "Taking limit is taking the standard part".

Answer 6

I would like to respond more directly to the following comment of the OP:

Now here is were my doubt starts! In the final step we put the value of h=0 say in that it is infinitesimal quantity.

I would say that the point is not to set $h$ equal to zero, but rather to discard it at the end of the calculation. Thus, the limit can be defined as follows: (1) one starts with an infinitesimal increment, and then (2) one "rounds off the value of the function to the nearest real number". Fermat, Leibniz, and others always chose their language carefully so as not to imply that one sets the increment equal to zero. Some relevant literature can be found here.