## Contents

# Limits of Functions

by: Michael Yeh, proud Member of the Math Squad.

keyword: tutorial, limit, function, sequence

## Introduction

Provided here is a brief introduction to the concept of "limit," which features prominently in calculus. To help motivate the definition, we first consider continuity at a point. Unless otherwise mentioned, all functions here will have domain and range $ \mathbb{R} $, the real numbers. Words such as "all," "every," "each," "some," and "there is/are" are quite important here; read carefully!

## Continuity at a point

Let's consider the following three functions along with their graphs (in blue). The red dots in each correspond to $ x=0 $, e.g. for $ f_1 $, the red dot is the point $ (0,f_1(0))=(0,0) $. Ignore the red dashed lines for now; we will explain them later.

- $ \displaystyle f_1(x)=x^3 $

- $ f_2(x)=\begin{cases}-x^2-\frac{1}{2} &\text{if}~x<0\\ x^2+\frac{1}{2} &\text{if}~x\geq 0\end{cases} $

- $ f_3(x)=\begin{cases} \sin\left(\frac{1}{x}\right) &\text{if}~x\neq 0\\ 0 &\text{if}~x=0\end{cases} $

We can see from the graphs that $ f_1 $ is "continuous" at $ 0 $, and that $ f_2 $ and $ f_3 $ are "discontinuous" at 0. But, what exactly do we mean? Intuitively, $ f_1 $ seems to be continuous at $ 0 $ because $ f_1(x) $ is close to $ f_1(0) $ whenever $ x $ is close to $ 0 $. On the other hand, $ f_2 $ appears to be discontinuous at $ 0 $ because there are points $ x $ which are close to $ 0 $ but such that $ f_2(x) $ is far away from $ f_2(0) $. The same observation applies to $ f_3 $.

Let's make these observations more precise. First, we will try to estimate $ f_1(0) $ with error at most $ 0.25 $, say. In the graph of $ f_1 $, we have marked off a band of width $ 0.5 $ about $ f_1(0) $. So, any point in the band will provide a good approximation here. As a first try, we might think that if $ x $ is close enough to $ 0 $, then $ f_1(x) $ will be a good estimate of $ f_1(0) $. Indeed, we see from the graph that for *any* $ x $ in the interval $ (-\sqrt[3]{0.25},\sqrt[3]{0.25}) $, $ f_1(x) $ lies in the band (or if we wish to be more pedantic, we would say that $ (x,f_1(x)) $ lies in the band). So, "close enough to $ 0 $" here means in the interval $ (-\sqrt[3]{0.25},\sqrt[3]{0.25}) $; note that *any* point which is close enough to $ 0 $ provides a good approximation of $ f_1(0) $.

There is nothing special about our error bound $ 0.25 $. Choose a positive number $ \varepsilon $, and suppose we would like to estimate $ f_1(0) $ with error at most $ \varepsilon $. Then, as above, we can find some interval $ \displaystyle(-\delta,\delta) $ about $ 0 $ (if you like to be concrete, any $ \displaystyle\delta $ such that $ 0<\delta<\sqrt[3]{\varepsilon} $ will do) so that for *any* $ x $ in $ \displaystyle(-\delta,\delta) $, $ f_1(x) $ will be a good estimate for $ f_1(0) $, i.e. $ f_1(x) $ will be no more than $ \varepsilon $ away from $ f_1(0) $.

Can we do the same for $ f_2 $? That is, if $ x $ is close enough to $ 0 $, then will $ f_2(x) $ be a good estimate of $ f_2(0) $? Well, we see from the graph that $ f_2(0.25) $ provides a good approximation to $ f_2(0) $. But if $ 0.25 $ is close enough to $ 0 $, then certainly $ -0.25 $ should be too; however, the graph shows that $ f_2(-0.25) $ is not a good estimate of $ f_2(0) $. In fact, for any $ x>0 $, $ f_2(-x) $ will never be a good approximation for $ f_2(0) $, even though $ x $ and $ -x $ are the same distance from $ 0 $.

In contrast to $ f_1 $, we see that for any interval $ \displaystyle(-\delta,\delta) $ about $ 0 $, we can find an $ x $ in $ \displaystyle(-\delta,\delta) $ such that $ f_2(x) $ is more than $ 0.25 $ away from $ f_2(0) $.

The same is true for $ f_3 $. Whenever we find an $ x $ such that $ f_3(x) $ lies in the band, we can always find a point $ y $ such that 1) $ y $ is just as close or closer to $ 0 $ and 2) $ f_3(y) $ lies outside the band. So, it is not true that if $ x $ is close enough to $ 0 $, then $ f_3(x) $ will be a good estimate for $ f_3(0) $.

Let's summarize what we have found. For $ f_1 $, we saw that for each $ \varepsilon>0 $, we can find an interval $ \displaystyle(-\delta,\delta) $ about $ 0 $ ($ \displaystyle\delta $ depends on $ \varepsilon $) so that for every $ x $ in $ \displaystyle(-\delta,\delta) $, $ |f_1(x)-f_1(0)|<\varepsilon $. However, $ f_2 $ does not satisfy this property. More specifically, there is an $ \varepsilon>0 $, namely $ \varepsilon=0.25 $, so that for any interval $ \displaystyle(-\delta,\delta) $ about $ 0 $, we can find an $ x $ in $ \displaystyle(-\delta,\delta) $ such that $ |f_2(x)-f_2(0)|\geq\varepsilon $. The same is true of $ f_3 $.

Now we state the formal definition of continuity at a point. Compare this carefully with the previous paragraph.

**DEFINITION 1.** Let $ f $ be a function from $ \displaystyle A $ to $ \mathbb{R} $, where $ A\subset\mathbb{R} $. Then $ f $ is continuous at a point $ c\in A $ if for every $ \varepsilon>0 $, there is a $ \displaystyle\delta>0 $ such that $ |f(x)-f(c)|<\varepsilon $ for any $ x $ that satisfies $ \displaystyle|x-c|<\delta $. $ f $ is said to be continuous if it is continuous at every point of $ A $.

In our language above, $ \varepsilon $ is the error bound, and $ \displaystyle\delta $ is our measure of "close enough (to $ c $)." Note that continuity is defined only for points in a function's domain. So, the function $ k(x)=1/x $ is technically continuous because $ 0 $ is not in the domain of $ k $. If, however, we defined $ k(0)=0 $, then $ k $ will no longer be continuous.

## The Limit of a Function at a Point

Now, let's consider the two functions $ g_1 $ and $ g_2 $ below. Note that $ g_1 $ is left undefined at $ 0 $.

- $ g_1(x)=\begin{cases}-x^2-x &\text{if}~x<0\\ x&\text{if}~x>0\end{cases} $

- $ g_2(x)=\begin{cases}0 &\text{if}~x\neq 0\\ \frac{1}{2}&\text{if}~x=0\end{cases} $

Recall the function $ f_1 $ from the previous section. We found that it was continuous at $ 0 $ because $ f_1(x) $ is close to $ f_1(0) $ if $ x $ is close enough to $ 0 $. We can do something similar with $ g_1 $ and $ g_2 $ here. From the graph, we can see that $ g_1(x) $ is close to $ 0 $ whenever $ x $ is close enough, but not equal, to $ 0 $. Similarly, we see that $ g_2(x) $ is close to $ 0 $ whenever $ x $ is close enough, but not equal, to $ 0 $. The "not equal" part is important for both $ g_1 $ and $ g_2 $ because $ g_1 $ is undefined at $ x=0 $ while $ g_2 $ has a discontinuity there. The idea is similar to that of continuity, but we ignore whatever happens at $ x=0 $. We are concerned more with how $ g_1 $ and $ g_2 $ behave around $ 0 $ rather than at $ 0 $. This leads to the following definition.

**DEFINITION 2.** Let $ f $ be a function defined for all real numbers, with possibly finitely many exceptions, and with range $ \mathbb{R} $. Let $ c $ be any real number. We say that the limit of $ f $ at $ c $ is $ b $, or that the limit of $ f(x) $ as $ x $ approaches $ c $ is $ b $, and write $ \lim_{x\to c}f(x)=b $ if for every $ \varepsilon>0 $, there is a $ \displaystyle\delta>0 $ such that $ |f(x)-b|<\varepsilon $ whenever $ \displaystyle 0<|x-c|<\delta $.

This is the same as the definition for continuity, except we ignore what happens at $ c $. We can see this in two places in the above definition. The first is the use of $ b $ instead of $ f(c) $, and the second is the condition $ \displaystyle 0<|x-c|<\delta $, which says that $ x $ is close enough, but not equal, to $ c $. The restriction on the domain of $ f $ in the above definition is not really necessary; if $ f $ has domain $ A\subset\mathbb{R} $, we can define $ \lim_{x\to c}f(x) $ for any $ c $ which is a "limit point" of $ \displaystyle A $. You can consult Rudin's *Principles of Mathematical Analysis,* for example, if you're interested. But, in most calculus courses, one usually encounters the situation described in the definition.

If you've ever seen *Mean Girls,* you know that the limit does not always exist. For example, if $ f_2 $ is the function in the previous section, then $ \lim_{x\to 0}f_2(x) $ does not exist. We will show this rigorously in the exercises, but it is clear from the graph of $ f_2 $.

Having defined what a limit is, we make a few remarks about how continuity and limit are related. As a first observation, we can now restate the definition of continuity at a point more succinctly in terms of limits: $ f $ is continuous at $ c $ if and only if $ \displaystyle\lim_{x\to c}f(x)=f(c) $. Now, let's take a look at the functions $ g_1 $ and $ g_2 $ defined earlier. $ g_1 $ is continuous at every point of its domain; however, it is undefined at $ 0 $. But since $ \lim_{x\to 0}g_1(x) $ exists and equals $ 0 $, we can make $ g_1 $ continuous on the entire real line by defining $ g_1(0)=0 $. We note that even though $ \lim_{x\to 0}g_2(x) $ exists, $ g_2 $ is discontinuous at $ 0 $; this is because $ \lim_{x\to 0}g_2(x)=0\neq \frac{1}{2}=g_2(0) $. This shows that the existence of $ \displaystyle\lim_{x\to c}f(x) $ does not imply continuity at $ c $, even though, as we just mentioned, continuity of $ f $ at $ c $ implies the existence of $ \displaystyle\lim_{x\to c}f(x) $.

## Limits at Infinity

Let $ f $ be as in Definition 2. So far, we have only discussed $ \displaystyle\lim_{x\to c}f(x) $ when $ c $ is a real number. It is not difficult to modify Definition 2 for when $ c=\pm\infty $. In Definition 2, we considered $ x $ to be "close enough" when $ x $ is in the interval $ \displaystyle(c-\delta,c+\delta) $ and $ x\neq c $. For $ \infty $, the analogue would be that $ x $ belongs to an interval of the form $ \displaystyle(M,\infty) $, where $ M $ is a real number, and similarly for $ -\infty $. So, we say that the limit of $ f $ at $ \infty $ (respectively, $ -\infty $) is $ b $ and write $ \displaystyle\lim_{x\to\infty}f(x)=b $ (respectively, $ \displaystyle\lim_{x\to-\infty}f(x)=b $) if for every $ \varepsilon>0 $, there is a real number $ M $ so that $ |f(x)-b|<\varepsilon $ whenever $ x>M $ (respectively, $ x<M $).

## Exercises

The triangle inequality will be handy for a few of these.

A) Show that any constant function is continuous.

B) Suppose $ \lim_{x\to c}f(x)=a $ and $ \lim_{x\to c}g(x)=b $. Show that

- i)$ \lim_{x\to c}(f+g)(x)=a+b $,
- ii)$ \lim_{x\to c}(f-g)(x)=a-b $, and
- iii)$ \lim_{x\to c}(fg)(x)=ab $.

Furthermore, if $ b\neq0 $, show that

- iv)$ \lim_{x\to c}\left(\frac{f}{g}\right)(x)=\frac{a}{b} $.

As a consequence of A) and B), any polynomial function is continuous.

C) Show that $ \lim_{x\to 0}f_2(x) $ does not exist.

D) Compute the following limits (if they exist):

- i)$ \lim_{x\to0}7 $
- ii)$ \lim_{x\to2}5x^2-3 $
- iii)$ \lim_{x\to+\infty}\frac{2x^2-3x^2-1}{x^3+5} $
- iv)$ \lim_{x\to+\infty}\frac{x^3-10x^2+4}{4x^3-2} $
- v)$ \lim_{x\to-2}\frac{5x^2-3}{x+2} $

E) (Derivatives) Let $ f $ be a function from $ \mathbb{R} $ to $ \mathbb{R} $ and $ c\in\mathbb{R} $. If $ \lim_{h\to 0}\frac{f(c+h)-f(c)}{h} $ exists, we say that $ f $ is differentiable at $ c $. The above limit is then called the "derivative of $ f $ at $ c $" and is denoted by $ \displaystyle f'(c) $. Intuitively, $ f'(c) $ is the instantaneous rate of change of $ f $ at the point $ c $; it is also the slope of the line tangent to the graph of $ f $ at the point $ (c,f(c)) $.

- i) Let $ \displaystyle f(x)=x^2 $. Show that $ f $ is differentiable everywhere, and compute $ \displaystyle f' $.
- ii) Let $ \displaystyle g(x)=|x| $. Show that $ g $ is not differentiable at $ 0 $.

## Solutions to Exercises

A) Let $ f(x)=k $ be a constant function with value $ k $ and let $ c $ be a real number. Now, let $ \varepsilon>0 $ be given and set $ \displaystyle\delta=1 $. Then for *any* $ x $ in the interval $ \displaystyle(c-\delta,c+\delta) $, $ |f(x)-f(c)|=|k-k|=0<\varepsilon $.

B) Let $ \varepsilon>0 $ be given. Then we can find a $ \displaystyle\delta_1>0 $ so that $ |f(x)-a|<\varepsilon $ whenever $ \displaystyle0<|x-c|<\delta_1 $, and a $ \displaystyle\delta_2>0 $ so that $ 0<|g(x)-b|<\varepsilon $ whenever $ \displaystyle0<|x-c|<\delta_2 $. Set $ \displaystyle\delta=\min\{\delta_1,\delta_2\} $. Then, for any $ x $ such that $ \displaystyle0<|x-c|<\delta $,

- i)
- $ |(f+g)(x)-(a+b)|=|(f(x)-a)+(g(x)-b)|\leq|f(x)-a|+|g(x)-b|<\varepsilon+\varepsilon=2\varepsilon $.

Also, it doesn't really matter that we have $ 2\varepsilon $ on the right side instead of $ \varepsilon $ because $ 2\varepsilon $ can be made as small as we like. That is, we can think of $ 2\varepsilon $ as our error bound, instead of $ \varepsilon $.

- ii)
- $ |(f-g)(x)-(a-b)|=|(f(x)-a)+(b-g(x))|\leq|f(x)-a|+|b-g(x)|<2\varepsilon $.
- iii) Note that by the Triangle Inequality, $ |f(x)|-a\leq|f(x)-a|<\varepsilon $ so that $ |f(x)|<|a|+\varepsilon $. Then,
- $ \begin{align}|(fg)(x)-ab|&=|f(x)g(x)-f(x)b+f(x)b-ab|\\ &\leq|f(x)g(x)-f(x)b|+|f(x)b-ab|\\ &=|f(x)||g(x)-b|+|f(x)-a||b|\\ &< (|a|+\varepsilon)|g(x)-b|+|f(x)-a||b|\\ &<(|a|+\varepsilon)\varepsilon+\varepsilon|b|\\ &=(|a|+|b|+\varepsilon)\varepsilon\end{align} $

Again, we are fine since $ (|a|+|b|+\varepsilon)\varepsilon $ can be made as small as we like.

- iv) By part (iii), it will suffice to show that $ \lim_{x\to c}\left(\frac{1}{g}\right)(x)=\frac{1}{b} $. Let $ \varepsilon>0 $ and set $ \eta=\frac{1}{2}\min\{\varepsilon,|b|\} $. Note that $ \displaystyle\eta>0 $ since $ b\neq 0 $. Therefore, we can find a $ \displaystyle\delta>0 $ so that $ \displaystyle|g(x)-b|<\eta $ whenever $ \displaystyle 0<|x-c|<\delta $. So, if $ \displaystyle 0<|x-c|<\delta $, then $ |g(x)-b|<\varepsilon $ since $ \eta\leq\frac{1}{2}\varepsilon<\varepsilon $. Also, by the triangle inequality we have $ |b|-|g(x)|\leq|b-g(x)|<\eta $. But also, $ \eta\leq\frac{1}{2}b $, so we get $ |g(x)|\geq|b|-\eta\geq|b|-\frac{1}{2}\varepsilon=\frac{1}{2}|b| $. Putting these together, we find that for any $ x $ such that $ \displaystyle 0<|x-c|<\delta $,
- $ \bigg|\left(\frac{1}{g}\right)(x)-\frac{1}{b}\bigg|=\frac{|b-g(x)|}{|g(x)||b|}<\frac{\varepsilon}{\left(\frac{1}{2}|b|\right)|b|}=2\frac{\varepsilon}{|b|^2} $.

C) Suppose that the limit did exist, say $ \lim_{x\to 0}f_2(x)=b $. Then setting $ \varepsilon=\frac{1}{4} $, there is a $ \displaystyle\delta>0 $ so that $ |f_2(x)-b|<\varepsilon $ whenever $ \displaystyle 0<|x-0|<\delta $. In particular, $ |f_2(-\frac{1}{2}\delta)-b|<\varepsilon $ and $ |f_2(\frac{1}{2}\delta)-b|<\varepsilon $.

Since $ -\frac{1}{2}\delta<0 $, we have $ f_2(-\frac{1}{2}\delta)=-(-\frac{1}{2}\delta)^2-\frac{1}{2}=(\frac{1}{2}\delta)^2-\frac{1}{2} $, and since $ \frac{1}{2}\delta>0 $, we have $ f_2(\frac{1}{2}\delta)=(\frac{1}{2}\delta)^2+\frac{1}{2}=(\frac{1}{2}\delta)^2-\frac{1}{2} $.

So, $ |f_2(\frac{1}{2}\delta)-f_2(-\frac{1}{2}\delta)|=|2(\frac{1}{2}\delta)^2+1|=2(\frac{1}{2}\delta)^2+1>1 $.

Now, the triangle inequality gives

- $ 1<|f_2(\frac{1}{2}\delta)-f_2(-\frac{1}{2}\delta)|=|f_2(\frac{1}{2}\delta)-b+b-f_2(-\frac{1}{2}\delta)|\leq |f_2(\frac{1}{2}\delta)-b|+|b-f_2(-\frac{1}{2}\delta)|<2\varepsilon=\frac{1}{2} $.

But of course, $ 1\nless\frac{1}{2} $, so we obtain a contradiction. This shows that $ \lim_{x\to 0}f_2(x) $ does not exist.

D)

- i) $ \displaystyle 7 $
- ii) $ \displaystyle 17 $
- iii) $ \displaystyle 0 $
- iv) $ \frac{1}{4} $
- v) does not exist

E)

- i) Let $ h $ be a nonzero real number. Then for any $ c $,
- $ \frac{(c+h)^2-c^2}{h}=\frac{(c^2+2ch+h^2)-c^2}{h}=\frac{2ch+h^2}{h}=2c+h $.

The limit of the right hand side as $ h $ approaches $ 0 $ exists and equals $ 2c $ using A) and B). Thus, $ f $ is differentiable everywhere, and $ f'(c)=2c $.

- ii) Note that for nonzero $ h $,
- $ \frac{g(0+h)-g(0)}{h}=\frac{|0+h|-|0|}{h}=\frac{|h|}{h} $.

We notice that for $ h>0 $,

- $ \frac{g(0+h)-g(0)}{h}=\frac{|h|}{h}=\frac{h}{h}=1 $

while for $ h<0 $,

- $ \frac{g(0+h)-g(0)}{h}=\frac{|h|}{h}=\frac{-h}{h}=-1 $.

Then, as in C), $ \lim_{h\to 0}\frac{g(0+h)-g(0)}{h} $ does not exist. So, $ g $ is not differentiable at $ 0 $.

## Questions and comments

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