New Zealand
Level 8 - NCEA Level 3

# Introduction to Limits

Lesson

Many sequences display trends as they continue along - they may get closer to a single value, or become larger and larger. The same is true of many functions as their input value becomes large.
Additionally, some functions are undefined at particular input values - for example, the function $\frac{1}{x^2}$1x2 when $x=0$x=0, or $\tan x$tanx whenever $\cos x=0$cosx=0 - and the behaviour of such functions for inputs near these values is often quite interesting.

In order to talk about these trends - both when sequences or inputs get large, or when inputs get close to a breaking value - we use a concept called a limit. Luckily for us, we only need this single concept to discuss each of these scenarios in a consistent way.

Here's an example from ancient history:

Zeno was a Greek philosopher who lived in the period 490 to 430 BCE. Achilles was a very famous hero of Greek tradition, an athlete and warrior without equal. But Zeno wondered if he really was as fast as people said. Here's his paradox:

"Achilles and a tortoise are in a footrace. Achilles is feeling confident, so he gives the tortoise a $100$100 m head start. They begin, and Achilles runs $100$100 m to where the tortoise started - by the time Achilles gets there, the tortoise has moved $10$10 m ahead. Achilles moves that extra $10$10 m, but by then the tortoise has moved ahead again.

At each step of the process, Achilles is always behind the tortoise.

Whenever Achilles reaches the place where the tortoise was, there will always be more distance to go. If there is always more distance to go, then Achilles will never catch the tortoise."

Now we know that most people can run faster than a tortoise - you don't need to be a Greek demigod to catch one! But to properly explain it, we need to say that "the sum of distances between Achilles and the tortoise has a finite limit".

### Limits in sequences

Consider this process:

• Start with a number $x$x
• Add half the number $\left(\frac{x}{2}\right)$(x2) to it
• Then add half of that $\left(\frac{x}{2^2}\right)$(x22) to the result

...and so on. The terms obtained in this way can be written in the form of a sequence:

$\left\{x,\frac{3x}{2},\frac{7x}{4},\frac{15x}{8},\frac{31x}{16},\ldots\right\}${x,3x2,7x4,15x8,31x16,}

We can write an expression for the $n$nth term in the sequence, which will have the form

$\frac{\left(2^n-1\right)x}{2^{n-1}}$(2n1)x2n1.

By either considering the sequence or the general form of a term, we can see that the coefficient of $x$x in the numerator is becoming closer and closer to being double the value of the numerator - that is, the sequence is approaching the value $2x$2x. Now let's make two important observations:

1. As the process continues, the amount we are adding is decreasing. In fact, we are adding half the difference between the current term and $2x$2x, so we are going half of the rest of the way to $2x$2x each time. Therefore, the sequence will never exceed $2x$2x.
2. For any value less than $2x$2x, we can continue the process long enough to reach that value - in other words, we can get as close as we like to $2x$2x. For example, if we want to get closer to $2x$2x than $1.999x$1.999x, we need to reach the $11$11th term; if we want to get closer than $1.999$1.999$999x$999x, we need to reach the $21$21st term, and so on.

In light of these observations, we say the sequence written above converges to, or approaches, the value $2x$2x as the index $n$n becomes arbitrarily large. We express this in mathematical symbols as follows:

$\lim_{n\rightarrow\infty}\frac{\left(2^n-1\right)x}{2^{n-1}}=2x$limn(2n1)x2n1=2x

We then say that $2x$2x is the limit of the sequence.

Not all sequences have a finite limit. Consider the trending behaviour of this sequence

$\left\{x,2x,3x,4x,5x,\ldots\right\}${x,2x,3x,4x,5x,}

which gets larger and larger as it continues, giving rise to the expression

$\lim_{n\rightarrow\infty}nx=\infty$limnnx=.

In this case we say the sequence has a limit of infinity - it moves towards infinity, but (obviously!) never gets there.

### Long-term trends in functions and horizontal asymptotes

A function can behave in a few different ways as $x$x becomes very large - unlike sequences, we need to consider both the positive and the negative directions of "large". Some functions, like $f\left(x\right)=x^2$f(x)=x2, become larger and larger the further away the input is from the $y$y-axis. We can tell this from the graph of the function:

We write, using the same notation as before,

$\lim_{x\rightarrow\infty}x^2=\infty$limxx2= and $\lim_{x\rightarrow-\infty}x^2=\infty$limxx2=

to express that the function values become arbitrarily large and positive as the input values trend towards infinity.

Others, like $\frac{1}{x^2}$1x2, become smaller and smaller, trending towards a finite value - in this case $0$0. We can see this from its graph as well:

In both the negative and positive directions, the graph becomes closer to the $x$x-axis as $x$x increases.

We write

$\lim_{x\rightarrow\infty}\frac{1}{x^2}=0$limx1x2=0 and $\lim_{x\rightarrow-\infty}\frac{1}{x^2}=0$limx1x2=0

to express that the function values become arbitrarily close to $0$0 as the input values trend towards infinity. Importantly, this function looks more and more like the line $y=0$y=0 the further from the axis we travel. This line is a horizontal asymptote for the function, and is usually drawn as a dashed line.

Another example of a function with a horizontal asymptote is $f(x)=\frac{x^2}{x^2-1}$f(x)=x2x21.
If we put a very large number (either positive or negative) into this function, the answer will be very close to $1$1. In fact, the bigger the input we choose, the closer the output will be to $1$1. We therefore say that $y=1$y=1 is a horizontal asymptote, and write

$\lim_{x\rightarrow\infty}\frac{x^2}{x^2-1}=1$limxx2x21=1 and $\lim_{x\rightarrow-\infty}\frac{x^2}{x^2-1}=1$limxx2x21=1.

We draw in the horizontal asymptote on the graph as follows:

These two types of trends (to infinity, and to a finite value) can occur in the same function, depending on whether we are trending towards $\infty$ or $-\infty$, such as

$\lim_{x\rightarrow\infty}e^x=\infty$limxex= and $\lim_{x\rightarrow-\infty}e^x=0$limxex=0.

We still say that $f(x)=e^x$f(x)=ex has the horizontal asymptote $y=0$y=0 even though it only trends towards this line in one direction.

But what about a function like $f\left(x\right)=\sin x$f(x)=sinx? We know from the graph of $\sin x$sinx that its output is always between $1$1 and $-1$1, so the limit is not infinite in either direction. But it also never "settles down", trending closer and closer to a single value, in the same way as the other examples with finite limits we have seen already:

The function does not look more and more like any horizontal line as $x$x becomes large - it does not have an asymptote, and we say that the limit $\lim_{x\rightarrow\pm\infty}\sin x$limx±sinx does not exist. A similar idea is true for sequences - if the sequence oscillates between different values and never settles down, the limit does not exist in the same way.

### Undefined values and vertical asymptotes

We have seen limits of functions as the input value $x$x becomes very large. But the notion of a limit works just as well for finite values of $x$x. Most finite limits for most functions don't tell us any more than we already knew - for example, the limit

$\lim_{x\rightarrow2}$limx2  $x^2$x2

can be read as "the value of $x^2$x2 as  $x$x approaches $2$2". This is $2^2=4$22=4 - the value we get when we substitute $2$2 into $x^2$x2. The real power of limits is their ability to assign values to expressions like

$\lim_{x\rightarrow0}\frac{\sin x}{x}$limx0sinxx.

If we simply substitute $x=0$x=0 into $\frac{\sin x}{x}$sinxx we get $\frac{0}{0}$00, showing that the function $f\left(x\right)=\frac{\sin x}{x}$f(x)=sinxx is undefined for $x=0$x=0. However, examining a table of values for the function reveals that approaching $0$0 brings us closer and closer to a particular function value:

 $x$x $-0.1$−0.1 $-0.01$−0.01 $-0.001$−0.001 $0$0 $0.001$0.001 $0.01$0.01 $0.1$0.1 $\frac{\sin x}{x}$sinxx​ $0.99833$0.99833 $0.99998$0.99998 $0.99999$0.99999 $0.99999$0.99999 $0.99998$0.99998 $0.99833$0.99833

Whether we approach from the left (negative side) or the right (positive side), the value of $\frac{\sin x}{x}$sinxx approaches $1$1. This is further confirmed by looking at the graph for the function:

We therefore write

$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$limx0sinxx=1.

This does not say that $\frac{\sin0}{0}=1$sin00=1!  The function is still not defined for $x=0$x=0. But we can say that the function value becomes as close as we like to $1$1 by approaching $0$0.

Sometimes limits towards a finite value of $x$x can produce infinities. Let's reconsider the graph of $\frac{1}{x^2}$1x2:

In this example, as $x$x approaches $0$0, the value of the function becomes larger and larger. Let's move the function over a little to see this better; here is the graph of $\frac{1}{\left(x-1\right)^2}$1(x1)2:

Here the function value becomes larger and larger as  $x$x approaches $1$1. The dashed line is called a vertical asymptote. Just like the horizontal asymptote we saw earlier, this is a line that the function draws arbitrarily close to. In limit form we write

$\lim_{x\rightarrow0}\frac{1}{x^2}=\infty$limx01x2=, and $\lim_{x\rightarrow1}\frac{1}{(x-1)^2}=\infty$limx11(x1)2=.

There are a few things to keep in mind:

1. A function with an undefined value doesn't automatically have an asymptote. The function $f\left(x\right)=\frac{\sin x}{x}$f(x)=sinxx does  not have a vertical asymptote at $x=0$x=0, since the function doesn't look like a vertical line there.
2. A function can have an asymptote without having a limit, which occurs if the path of approach changes the behaviour of the function near the limiting value.

To illustrate the second point, consider the graph of $y=\frac{1}{x-1}$y=1x1 as $x$x approaches $1$1:

Notice how approaching the limiting value from the left produces numbers that are larger and more  negative, while approaching the limiting value from the right produces numbers that are larger and positive. The function has a vertical asymptote at $x=1$x=1, but

the limit $\lim_{x\rightarrow1}\frac{1}{x-1}$limx11x1 does not exist.

A similar thing is true for the function $f\left(x\right)=\tan x$f(x)=tanx. It has infinitely many $x$x-values for which the function is undefined, and infinitely many asymptotes, though the limit as $x$x approaches any of these values does not exist:

The function moves in opposite directions on either side of an asymptote line.

Each of these concepts will be explored in greater detail in the coming lessons. For now, let's summarise what we've seen in this chapter.

Introduction to Limits
1. Limits in sequences: We can discuss the trend of a sequence as the number of terms in the sequence becomes large. If the values of the sequence
• become arbitrarily large and positive (or negative), the limit is  $\infty$ (or $-\infty$),
• converge to a particular value, the limit is that value,
• oscillates between many values and never settles down, the limit does not exist.
2. Infinite limits in functions: We can discuss the trend of a function as its input becomes large (either positive or negative). If the function values
• become arbitrarily large and positive (or negative), the limit is  $\infty$ (or $-\infty$),
• converge to a particular value, the limit is that value. In this case there is a horizontal asymptote - a horizontal line that the function approaches arbitrarily closely,
• oscillates between many values and never settles down, the limit does not exist.
3. Finite limits in functions: We can discuss the trend of a function as the input approaches a finite value. If the function is defined for that input, then the limit is the function value of that input. If it is not defined for that input, then if approaching the input from both the negative and the positive directions
• produces the same finite function value, the limit is that function value.
• both produce arbitrarily large and positive (or negative) values, then the limit is $\infty$ (or $-\infty$).
• produces different function values, or different signs of  $\infty$, then the limit does not exist.
In the last two cases, there is a vertical asymptote - a vertical line that the function approaches arbitrarily closely.

#### Practice questions

##### Question 1

Consider the sequence $9,3,1,\frac{1}{3},\ldots$9,3,1,13,

1. What is the next term in the sequence?

2. What value do the terms approach as we progress along the sequence?

3. Will there eventually be a term in the sequence that is equal to zero?

No

A

Yes

B

No

A

Yes

B

##### Question 2

Consider the function $f\left(x\right)=\frac{1}{7-x}$f(x)=17x.

1. Complete the following table of values, in which $x<7$x<7.

 $x$x $5$5 $6$6 $6.9$6.9 $6.99$6.99 $f\left(x\right)$f(x) $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
2. Complete the following table of values, in which $x>7$x>7.

 $x$x $9$9 $8$8 $7.1$7.1 $7.01$7.01 $f\left(x\right)$f(x) $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
3. What is the limit of $f\left(x\right)$f(x) as the value of $x$x approaches $7$7?

The limit does not exist.

A

The limit is $0$0.

B

The limit is $\infty$.

C

The limit is $-\infty$.

D

The limit does not exist.

A

The limit is $0$0.

B

The limit is $\infty$.

C

The limit is $-\infty$.

D

##### Question 3

Consider the function that has been graphed below.

1. What value does $y$y approach as $x$x approaches infinity?

2. What value does $y$y approach as $x$x approaches negative infinity?

3. What value does $y$y approach as $x$x approaches zero?

### Outcomes

#### M8-10

Identify discontinuities and limits of functions

#### 91578

Apply differentiation methods in solving problems