Confusion with $ \epsilon$-$\delta $ definition.

Question

In delta-epsilon definition of limit what I understand is that for whatever small value of $ \epsilon $ is chosen if the corresponding $ \delta$ is found then the existence of the limit is confirmed. But the problem is that we are not sure how big the $ \delta $ is gonna be. So how does only the presence of $ \delta$ proves the existence of limit? Why the size of $ \delta$ doesn't matter? Because in case of almost horizontal curves you may get pretty big value for $ \delta $ even though the $ \epsilon $ is sufficiently small. Now all these questions actually arise from the layman's definition of the limit that is: if x goes more and more towards a specific value then the value towards which functional values move is the limit of that function at that specific point. So, in this definition we see input values x actually move more and more towards that specific point that is $ \delta$ gets smaller and smaller.

Answer 1

To answer your question, let's first look at an example.

Let $f(x) = 1$ for all $x$. What is $$\lim_{x \to 3} f(x)?$$

Obviously it is just $1$. Now look at a slightly less trivial example, say $$f(x) = \begin{cases}1, & x \ne 3 \\ 2, & x = 3. \end{cases}$$

Again, $\displaystyle \lim_{x \to 3} f(x) = 1$, although this time, the value of the function itself is not $1$ at $x = 3$.

How do we frame this in the context of the epsilon-delta definition of limit? It states that for any $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $0 < |x - 3| < \delta$, then $|f(x) - 1| < \epsilon$. But notice that $|x - 3| > 0$ implies that $x \ne 3$, hence $f(x) = 1$ for any such $x$, hence $|f(x) - 1| = 0$. This means that for any $\epsilon > 0$, the choice of $\delta$ is irrelevant: any $x \ne 3$ will make $|f(x) - 1| = 0 < \epsilon$.

Understanding this example is the beginning of the answer to your question. The quantity $\epsilon$ represents in a sense the maximum error tolerance between the function's value $f(x)$ and the limiting value $L$. That is why we must be free to choose to make it as small as we please, because it is the quantity that determines how "closely" the value of the function gets to the limiting value.

The quantity $\delta$, on the other hand, merely describes some neighborhood around the limit point that $x$ approaches, for which all $x$-values inside this (punctured) neighborhood are guaranteed to give a corresponding function value $f(x)$ that is within the desired error tolerance $\epsilon$. So if the function is very "flat" in such a neighborhood, it doesn't need to be small. In fact, with the example we discussed above, $f$ is constant except at the point of discontinuity, which is why any $\delta > 0$ is permissible.

Let's now turn our attention to a pair of examples. Consider $$f(x) = \begin{cases} \frac{x}{100}, & x \ne 3 \\ 0, & x = 3, \end{cases} \qquad g(x) = \begin{cases} 100x, & x \ne 3 \\ 0, & x = 3. \end{cases}$$ Let's compare $\displaystyle \lim_{x \to 3} f(x)$ versus $\displaystyle \lim_{x \to 3} g(x)$. Using the epsilon-delta definition on $f$, we require $|\frac{x}{100} - \frac{3}{100}| < \epsilon$ whenever $0 < |x - 3| < \delta$. So note that $$\left|\frac{x}{100} - \frac{3}{100}\right| = \frac{|x - 3|}{100},$$ hence the choice $$\delta = 100 \epsilon$$ assures us that whenever $|x-3| < \delta = 100 \epsilon$, then $$|f(x) - \frac{3}{100}| = \frac{|x-3|}{100} < \frac{\delta}{100} = \frac{100\epsilon}{100} = \epsilon.$$

Similarly, for $g(x)$, it is easy to see that the choice $$\delta = \frac{\epsilon}{100}$$ satisfies the conditions of the limit definition. The details are left as an exercise.

Now compare the size of these two intervals for the same choice of $\epsilon$. For instance, suppose we choose $\epsilon = 0.1$. That is to say, our error tolerance for $f(x)$ and $g(x)$ is only $0.1$. In the case of $f$, this means that all $x$ in a neighborhood of radius $\delta = 100(0.1) = 10$ will meet the desired error tolerance. In the case of $g$, however, the radius is much smaller: $\delta = \frac{1}{100}(0.1) = 0.001$.

Geometrically, we can understand why this is the case: the function $f$ has a much flatter slope than $g$, meaning that its values in a neighborhood around $x = 3$ do not vary as much compared to the much steeper slope we see in $g$. So in order to achieve the same error tolerance of $0.1$, the resulting interval around $x = 3$ must be much smaller for $g$ than for $f$. In each case, the limit still exists, but what this tells us is that the size of $\delta$ is not important, because as long as it guarantees that for all $x$-values in the resulting neighborhood, the function's values will fall within the error tolerance $\epsilon$, then it is sufficient to prove the limit exists.

Answer 2

The first reason why we want to be able to pick and choose our $\delta$ freely is precisely the fact that, as you mentioned, when the graph of a function is a straight line through the point, the line might pass through that point at a very shallow angle or a very steep angle. But the angle of the line should not determine whether the function has a limit at that point. We need a definition that allows us to use an arbitrarily small $\delta$ for any given $\varepsilon$ so that we can talk about the limits of functions whose graphs are arbitrarily steep.

Also, consider a function like $f(x) = 2$. Surely the limit of that function at any value of $x$ is $2$. But no matter what $\varepsilon>0$ you choose, all the values of that function are between $2 - \varepsilon$ and $2 + \varepsilon$. So there really is no way to say how large $\delta$ should be. ("Infinity" isn't a real number, so it's not a candidate for a value of $\delta$.)

If you consider functions whose graphs are not straight lines, you have functions where the vertical distance and horizontal distance have different relationships as you approach from either side of the limit point.

You also have functions that are very hard to invert. For example, consider the limit of $f(x) = x^5 + x + 1$ at $x = 2$. Obviously, looking at the graph, the limit is $35$. But can you write a formula that will tell me, for any $\varepsilon$, exactly what $\delta$ I should choose so that $f(x + \delta) = 35 + \varepsilon$? You can't write this formula using only functions that are taught in an ordinary high school.

But those are just the very, very simplest cases.

A major difference between what many beginners think of as a "limit" and what mathematicians think of as a limit is that the beginner only contemplates limits of continuous functions that are monotone on each side of the limit point, so that you can imagine sliding a movable point along the curve of the function's graph toward the limit point, and as the horizontal distance to the limit point decreases, so does the vertical distance, always.

The mathematician wants to be able to take limits of functions that are much wilder. For the mathematician, there are functions where you can say (for example) that $$ \lim_{x\to2} f(x) = 5 $$ even though $f(x)$ is discontinuous at every value of $x$ except $x = 2$. For example, consider the function $$ f(x) = \begin{cases} x + 3 & \text{$x$ is rational}, \\ 5 & \text{$x$ is irrational} . \end{cases} $$

If you plot the function, it will have a dense set of points along the line $y = x + 3$ and a dense set of points along the line $y = 5$. It's hard to say what $y=f(x)$ does as $x$ approaches $2$ because in any interval of $x$ you have infinitely many values on the line $y = x + 3$ and infinitely many on the line $y = 5$. But if you zoom in on the graph toward the point $(x,y) = (2,5)$, it's clear that the points on the graph of the function all trend toward that point even though they come at it from four directions (left and right on two different lines) instead of just two directions on a single curve. We really want to say that $\lim_{x\to2} f(x) = 5$, and the delta-epsilon definition lets us do that.