The Concept ``Limit"

A long, long time ago, in a university far far away, a student doing his masters degree oral examination was asked ``If you had to sum up in one word what calculus was all about, what would it be ?". The student proceeded by cramming as much material as possible in a two minute pep talk. Then the examiner responded ``Actually, I was only looking for the word limit". In case you're wondering, the student did get his degree.

Introduction 1.1.1  
It is reasonable to claim that the foundation for all of calculus is indeed the concept ``limit". In this section we are going to explore what exactly is meant by

$$\lim_{x \to c} f(x) = L$$ (1.1.1)

where $f$ is a function, $c$ is a point and $L$ the value of this limit. First we will develop the idea intuitively and then we will work towards a formal definition of (1.1.1).
It is very tempting to simply interpret (1.1.1) as ``the functional values $f(x)$ can be made arbitrarily close to a unique number $L$ by choosing $x$ close enough to $c$ (but not equal to $c$)." In fact, in quite a few cases, this interpretation is all that you will need to understand what is going on. Sometimes, however, it is just not good enough; the weakness of it all is in the phrase ``by choosing $x$ close enough to $c$". In the following examples, we will see that it is not always obvious to interpret ``choosing $x$ close enough to $c$" in a correct and unambiguous way.

Example 1.1.2  
The graph of the function

$$f: \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto \frac{1}{x^4}$$ (1.1.2)

is as follows:

200pt

From this graph it is easy to see that if one approaches 0, $f(x)$ tails off to infinity. So in this case (1.1.1) would still make sense but with $L=\infty$. Note that $\infty$ is not a number but merely a symbol to express the unbounded behaviour of a function.

Example 1.1.3  
Look at the function

$$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: \... ... \mapsto x & \text{if $x\leq4$} \\ x \mapsto 2x-2 & \text{if $x>4$} \end{cases}$$ (1.1.3)

Its graph is as follows:

200pt

At the point $4$, there is no danger of the value of $f$ becoming infinite, but taking an $x$ and going closer and closer the $x$, one notices that if coming from the left $f(x)$ approaches $4$. Coming from the right, $f(x)$ approaches $6$.

Remark 1.1.4  
In example (1.1.3) the limit seemed to depend on the fact if we where coming from the left or from the right. This is clearly not the same situation as in example (1.1.2), where the limit bluntly tails off to infinity. In order to express the fact that there is something like a limit when coming from the left or from the right the terminology ``left hand limit" and ``right hand limit" is used. The common notation is

$$\lim_{x \to c^-} f(x) = L$$ (1.1.4)

for the left hand limit, i.e. going to $c$ through smaller numbers, and

$$\lim_{x \to c^+} f(x) = L$$ (1.1.5)

for the right hand limit, i.e. going to $c$ through bigger numbers.
In example (1.1.3), these would be

$$\lim_{x \to 4^-} f(x) = 4 \hbox{\ \ and\ \ }\lim_{x \to 4^+} f(x) = 6$$ (1.1.6)

Example 1.1.5  
In some cases it doesn't even matter if you come from the right or the left. The function behaves so wildly that different ways of going to a point give you different results! Take e.g. a look at the function

$$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: \... ...\frac{1}{x}) & \text{if $x\not=0$} \\ x \mapsto 0 & \text{if $x=0$} \end{cases}$$ (1.1.7)

The graph of this function looks like:

200pt

and is, quite frankly, not all that helpful in investigating the behaviour of $f(x)$ for $x$ going to 0. In any case, the graph points out one thing quite successfully, namely the extremely wild behaviour of $f$ close to 0. Take $n \in \hbox {I \hskip -5.2pt {N}}$ getting bigger, then

$$a_n := \frac{1}{2n\pi} \longrightarrow 0 \hbox{\ \ and\ \ } b_n := \frac{1}{(2n+1)\pi} \longrightarrow 0$$ (1.1.8)

So both are going to 0. On the other hand, we have

$$f(a_n) = \cos(2n\pi) = 1 \hbox{\ \ and\ \ } f(b_n) = \cos((2n+1)\pi)) = -1$$ (1.1.9)

Meaning that $f(a_n) \to 1$ and $f(b_n) \to -1$. So $\lim_{x \to 0} f(x)$ is not defined in this case, as both $1$ and $-1$ are possible outcomes depending through which points you go to get to 0. In fact any number of $[-1,1]$ can be attained by choosing a specific list of $a_n$ 's. That fact is not important: as soon as there is more than one single possible outcome, there is no limit. In order for the limit to exist, it should not matter at all through which points one goes.

Example 1.1.6  
Of course, sometimes the limit does exist. Take e.g. a look at the function

$$f: \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto \frac{\sin(x)}{x}$$ (1.1.10)

The graph of this function indicates that the limit in 0 is $1$, and indeed that is the case.

200pt

Looking at the expression $\frac{\sin(x)}{x}$, this is not that obvious, as $\frac{0}{0}$ is not defined. In fact $\frac{0}{0}$ can turn out to be 0, $\infty$ or any real number, as we will see later.

Definition 1.1.7  
The limit of a function $f$ at a point $c$ has value $L$, written as

$$\lim_{x \to c} f(x) = L \hbox{\ \ iff \ \ } \forall x : \forall \... ...h that } \vert x - c \vert < \delta \implies \vert f(x) - L \vert < \varepsilon$$ (1.1.11)

Remark 1.1.8  
Proving that a limit is indeed the value that you expect it to be using this $\varepsilon\hbox{-}\delta$ definition is somewhat cumbersome, yet necessary. This is not an exercise in pedantry, but a way to avoid falling into sometimes rather enticing traps like (1.1.7) where any number between $-1$ and $1$ seems like a valid outcome. Usually, it is relatively easy to guess what a limit will be and then we can prove that our ''guess" is correct by going through an $\varepsilon\hbox{-}\delta$ proof.

Example 1.1.9  
To prove that

$$\lim_{x \to 3} 2x+1 = 7$$ (1.1.12)

we have to go through an $\varepsilon\hbox{-}\delta$ procedure to be formally correct. First, fix an $\varepsilon$. Now, we have to find a $\delta > 0$ such that $\vert x-3\vert<\delta$ implies $\vert 2x+1-7\vert<\varepsilon$. To do this, note

$$\vert 2x+1-7\vert = \vert 2x-6\vert = 2 \vert x-3\vert < 2 \delta$$ (1.1.13)

So choosing $\delta < \frac{\varepsilon}{2}$ implies that $\vert 2x+1-7\vert<\varepsilon$ and we have proven that the limit exists and is indeed $7$.

Example 1.1.10  
In order to prove

$$\lim_{x \to 2} x^2+3x+1 = 11$$ (1.1.14)

we have to go through an $\varepsilon\hbox{-}\delta$ procedure, so first, fix an $\varepsilon$. Now, we have to find a $\delta > 0$ such that $\vert x-2\vert<\delta$ implies $\vert x^2+3x+1-11\vert<\varepsilon$. To do this, note

$$\vert x^2 + 3x + 1 - 11 \vert = \vert x^2 + 3x - 10 \vert \cr$$ (1.1.15)

The main idea in this calculation is to get $\vert x-2\vert$ in there somewhere, because that is how we can introduce our $\delta$. Now choose $\delta <1$ in order to get $\vert x-2\vert<1$, which implies $-1<x-2<1$ and therefore $1<x<3$. If $x<3$, then $\vert x+5\vert<8$, so we get

$$\vert x^2 + 3x + 1 - 11 \vert < \delta \cdot 8$$ (1.1.16)

So choosing $\delta < \min (1,\frac{\varepsilon}{8})$ implies that $\vert x^2+3x+1-11\vert<\varepsilon$ and we have proven that the limit exists and is indeed $11$.

Example 1.1.11  
Sometimes, a limit does not exist, e.g. there is no real number $L$ such that

$$\lim_{x \to 0} \frac{1}{x} = L$$ (1.1.17)

To see this, let $L$ be any number and take an $\varepsilon > 0$. We have

$$\left\vert \frac{1}{x} - L \right\vert < \varepsilon$$ (1.1.18)

which we can rewrite as

$$-\varepsilon < \frac{1}{x} - L < \varepsilon$$ (1.1.19)

and thus

$$L -\varepsilon < \frac{1}{x} < L + \varepsilon$$ (1.1.20)

Choosing $\varepsilon = 1$ (not exactly a small choice of $\varepsilon$), we get

$$\left\vert \frac{1}{x} \right\vert < \vert L \vert + 1$$ (1.1.21)

or written differently:

$$\left\vert x \right\vert > \frac{1}{\vert L\vert + 1}$$ (1.1.22)

For whatever value of $L$ that you choose, $x$ cannot be close to 0 Therefore, (1.1.18) cannot be right and therefore $\frac{1}{x}$ has no limit in 0.

Properties 1.1.12  
Let $c$ be any real number and take $f$ and $g$ so that they have limits at c. Furthermore take $k$ to be a constant. Then we have the following:

• Constant Rule
  $$\lim_{x \to c}k = k$$
• Limit of $x$ Rule
  $$\lim_{x \to c}x = c$$
• Multiple Rule
  $$\lim_{x \to c}kf(x) = k \lim_{x \to c}f(x)$$
• Sum Rule
  $$\lim_{x \to c}f(x) + g(x) = \lim_{x \to c}f(x) + \lim_{x \to c}g(x)$$
• Difference Rule
  $$\lim_{x \to c}f(x) - g(x) = \lim_{x \to c}f(x) - \lim_{x \to c}g(x)$$
• Product Rule
  $$\lim_{x \to c}f(x) g(x) = \lim_{x \to c}f(x) \lim_{x \to c}g(x)$$
• Quotient Rule
  $$\lim_{x \to c}\frac{f(x)}{g(x)} = \displaystyle\frac{\lim_{x \to c}f(x)}{\lim_{x \to c}g(x)}$$ if $$\lim_{x \to c}g(x) \not= 0$$
• Power Rule
  $$\lim_{x \to c}[f(x)]^\alpha = \left[\lim_{x \to c}f(x)\right]^\alpha$$
  where $\alpha$ can be any real number. If $\alpha$ is negative, then the formula is only valid if $\lim_{x \to c}f(x) \not=0$, to avoid division by 0.

Now, strictly speaking, we would have to prove each of the formulas above with an $\varepsilon\hbox{-}\delta$ argument, but this would take up too much time and too much paper. So we leave it to the extremely interested reader as an exercise.
These properties will turn out to be very useful when determining if a function is continuous or not

Lemma 1.1.13  
If $f(x) \leq g(x) \leq h(x)$ on a open interval around $c$ and if

$$\lim_{x \to c}f(x) = \lim_{x \to c}h(x) = L \hbox{\ then \ } \lim_{x \to c}g(x)= L$$ (1.1.23)

This lemma is often referred to as the ``Squeeze Rule" or the ``Sandwich Rule". It is mostly used with either $f$ or $h$ a constant function. It also exists for sequences (see Lemma 2.0.17) where the proof will be given.

Remark 1.1.14  
It is in the very nature of the definition of limit, that the differences between single variable calculus and multi variable calculus have their origin. Taking the limit in single variable calculus, you are essentially walking along an axis, so you have two possible directions: you either come from the left, or you come from the right. In multivariable calculus, you are at least walking on a plane and you therefore have an infinite number of directions you can choose to approach a certain point.
Take e.g. the function

$$f : \hbox {I \hskip -5.2pt {R}}^2 \to \hbox {I \hskip -5.2pt {R}}: (x,y) \mapsto \frac{xy^2}{x^2+y^4}$$ (1.1.24)

The graph of this function is

200pt

Going to $(0,0)$ can be done in a lot of ways. One way to go there would be along a line, The equation of any line going through $(0,0)$ is $y=mx$ where $m$ is the slope. On this line we have

$$f(x,y)=f(x,mx) =\frac{x(mx)^2}{x^2+(mx)^2} =\frac{m^2x^3}{x^2+m^4x^4} =\frac{m^2x}{1+m^4x^2}$$ (1.1.25)

Therefore $f(x,y) \to 0$ when $(x,y) \to (0,0)$ along a line. Any line. So the limit of $f(x,y)$ for $(x,y)$ going to $(0,0)$ should be 0 right ? Well, not quite. When going to $(0,0)$ while following the parabola $x=y^2$, we get

$$f(x,y)=f(y^2,y)=\frac{y^2 \cdot y^2}{(y^2)^2+y^4}=\frac{y^4}{2y^4}=\frac{1}{2}$$ (1.1.26)

so $f(x,y) \to \frac{1}{2}$ when $(x,y) \to (0,0)$ along the parabola $x=y^2$. Hence, $\lim_{(x,y) \to (0,0)} f(x,y)$ does not exist!