Limits of functions

The limit of a function is the value that the fuction approaches as the input approaches some value. Let's consider a function $f : \R\to\R$ and let $x_0\in\R$, the function will have a specific value at $x_0$, $f(x_0)$, but we will be interested in the values that $f$ takes close to $x_0$. In the next animation, which do you think the limit of the function $f$ is as $x$ tends to $x_0$?.

That definition of limit was a bit vague. Let's see the real mathematical definition.

Definition

Let $f$ be a function defined in all points inside some open interval $(a,b)$ that contains $x_0$, except perhaps in $x_0$. We say that the limit of $f$ as $x\to x_0$ ($x$ tends to $x_0$) is $L$ if for every $\varepsilon>0$ there exists $\delta>0$ (depending on $\varepsilon$) such that if $0<|x-x_0|<\delta$, then $|f(x)-L|<\varepsilon$. When this happens we write$$\lim_{x\to x_0} f(x) = L.$$

Test the definition in the next animation! We will calculate the limit of $f(x) = x-1$ as $x\to$$3$. By moving the value of $\varepsilon$$>0$, $\delta$ will be determined so that the conditions in the definition above are satisfied.

So what's all the fuss about? This seems trivial! Well, note that if the function had another definition at $x_0$, the limit would still be the same!

Limits don't "see" the value of the function at the point. It may not even be defined at $x_0$!

Proving a limit by definition

Now it's time we work on a limit by definition. Let's prove that thelimit of the following function$$f(x) = x^2+3,$$when $x\to1$, is $4$. Looking at the plot, this seems true, but we have to prove it!

The definition says that for any $\varepsilon>0$ we need to find $\delta>0$ such that some condition is satisfied. For each $\varepsilon$ we'll get a different $\delta$, so the idea will be to consider a generic $\varepsilon$ and see if we can define $\delta$ in terms of that $\varepsilon$. That way, whenever we take a specific $\varepsilon$, we can find $\delta$ immedaitely. Let's start!

Let $\varepsilon$$>0$ (this is the way of defining an arbitrary $\varepsilon$). We want to find $\delta$$>0$ such that if $0<|x-$$1$$|<$$\delta$, then $|f(x)-$$L$$|<$$\varepsilon$. We will start by trying to make $0<|x-$$1$$|$ appear from the equation $|f(x)-$$L$$|<$$\varepsilon$, so that then we can choose $\delta$ so that everything works. This won't be the proof, this is just the informal calculations, once we finish with them, we'll write the formal proof.

Replacing the values of $f$ and $L$ gives$$|f(x) - L| = |x^2 + 3 - 4| = |x^2 - 1| = |x-1||x+1|.$$We have $|x-$$1$$|$, which is exactly what we wanted. But we cannot yet replace it to make it smaller than $\delta$, because there is still dependence on $x$ in the term $|x+1|$. So we first need to get rid of it by bounding it from above. The idea is to take $\delta$ small, we can take it smaller than $1.$ Then if $|x-$$1$$|<$$\delta$$<1$, we have $-1<x-1<1$, so $1<x+1<3$. Great! Let's use this in the previous equation.$$|f(x) - L| = |x-1||x+1| < 3|x-1| < 3\delta.$$We want to have $|f(x)-$$L$$|<$$\varepsilon$ Then it suffices to take $3$$\delta$$<$$\varepsilon$, that is $\delta$$<$$\varepsilon$$/3.$ Recall that we also needed $\delta$$< 1,$ so we'll have to take the smaller of these two values.

This doesn't look like a formal proof! The actual proof is much shorter, once we know what $\delta$ to take.

Writing the proof

Let $\varepsilon$, choose $\delta$$< \min\{ 1,$$~ \varepsilon$$/3 \}.$ Then if $|x-$$1$$|<$$\delta$$$|f(x) - L| = |x-1||x+1| < 3|x-1| < 3\delta < \varepsilon.$$As we wanted to prove.

Properties of limits

I'm sure that, like me, you probably don't want to calculate limits by definition (like we just did) for the rest of your life. Luckily limits satisfy a list of properties that make them easier to handle! Let's learn them. From now on $f$ and $g$ are two functions and $c$ is a constant.