9.02 Introduction to limits

To move forward and calculate the gradient of a tangent to a function rather than estimating we need an understanding of limits and how to evaluate them.

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, small or approaches key points.

In order to talk about these trends we will introduce the concept of a limit.

Worked examples

Example 1

Find the limit of the function $f(x)=x^2$f(x)=x2 as $x$x approaches $3$3. That is, find $\lim_{x\rightarrow3}x^2$limx→3x2.

 

 

$x$x	$f(x)=x^2$f(x)=x2
$2.9$2.9	$8.41$8.41
$2.99$2.99	$8.9401$8.9401
$2.999$2.999	$8.994001$8.994001
$3.001$3.001	$9.006001$9.006001
$3.01$3.01	$9.0601$9.0601
$3.1$3.1	$9.61$9.61

 

 

 

 

 

 

 

 

From the graph and table above, we can observe that as $x$x approaches $3$3, the function $f(x)=x^2$f(x)=x2 approaches $9$9.  Approaching the value $x=3$x=3 from either direction results in the same limiting value of $9$9, which is also the value of the function at $x=3$x=3. For many functions such as this, the limit at a particular point is simply the function evaluated at that point. However, some functions are not so well-behaved.

Example 2

Find $\lim_{x\rightarrow0}\frac{\sin x}{x}$limx→0sinxx​.

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 from above or below 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	undefined	$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:

Therefore, $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$limx→0sinxx​=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 choosing an $x$x-value close enough to $0$0.

Example 3

Does the limit $\lim_{x\rightarrow2}f(x)$limx→2f(x), exist for the function shown in the graph below?

The limit $\lim_{x\rightarrow2}f(x)$limx→2f(x) does not exist, as the limit taken by approaching $2$2 from below does not agree with the limit if we approach $2$2 from above (often referred to as left-hand limit and right-hand limit). 

$\lim_{x\rightarrow2^-}f(x)=3$limx→2−f(x)=3 and $\lim_{x\rightarrow2^+}f(x)=5$limx→2+f(x)=5. Hence, we say$\lim_{x\rightarrow2}f(x)$limx→2f(x) is undefined.

 

• A limit fails to exist at a point if different values are reached when approaching the point from below or from above. We can, instead, speak of a left-hand limit as $x\rightarrow a^-$x→a− or a right-hand limit as $x\rightarrow a^+$x→a+.
• A limit at a point $x=a$x=a can fail to exist if the function values fluctuate rapidly near that point. You could check with a graphing calculator or other software the behaviour of the function defined by $f(x)=\sin\frac{1}{x}$f(x)=sin1x​ near $x=0$x=0. In this case, choose numbers for $x$x approaching $0$0 and the value of the function will not settle to a limit but vary rapidly between the extremes of $-1$−1 and $1$1. 
   

Example 4

Find $\lim_{x\rightarrow2}\frac{4x^2-8x}{x-2}$limx→24x2−8xx−2​.

Notice this graph is not defined at $x=2$x=2, so we cannot simply substitute in the value. Rather than looking at a graph or table we can simplify this function by factorising the top and dividing by the numerator. That is:

$\frac{4x^2-8x}{x-2}$4x2−8xx−2​	$=$=	$\frac{4x\left(x-2\right)}{x-2}$4x(x−2)x−2​
	$=$=	$4x$4x, for $x\ne2$x≠2

Thus for this function, everywhere except $x=2$x=2, the function will have the value $4x$4x. This means approaching from either side will result in approaching $4x$4x evaluated at $x=2$x=2, which is the limit $8$8. We can write:

$\lim_{x\rightarrow2}\frac{4x^2-8x}{x-2}$limx→24x2−8xx−2​	$=$=	$\lim_{x\rightarrow2}\frac{4x\left(x-2\right)}{x-2}$limx→24x(x−2)x−2​
	$=$=	$\lim_{x\rightarrow2}4x$limx→24x
	$=$=	$8$8

Notice the $\lim_{x\rightarrow2}$limx→2 remains in the expression until the limit is "taken" by substituting in the value $2$2. This technique of simplifying a rational expression to reveal the behaviour of the graph excluding the point of discontinuity is very important and we will make wide use of it in calculus.

The graph of this function will look like the graph of $y=4x$y=4x but with a single missing point at $x=2$x=2. As seen below.

 

Further behaviours at limits

Let's look briefly at a few different limiting behaviours you may come across.

Worked examples

Example 5

Describe the behaviour of $f(x)=x^2$f(x)=x2 as $x$x becomes a very large positive or negative number using limit notation.

The function $f\left(x\right)=x^2$f(x)=x2 becomes 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 can describe this behaviour using limit notation as follows:

$\lim_{x\rightarrow\infty}x^2=\infty$limx→∞x2=∞ and $\lim_{x\rightarrow-\infty}x^2=\infty$limx→−∞x2=∞

 

Example 6

Describe the behaviour of $f(x)=\frac{x^2}{x^2-1}$f(x)=x2x2−1​ as $x$x becomes a very large positive or negative number using limit notation. 

We can see the graph of the function is approaching a horizontal asymptote as $x$x approaches positive or negative infinity.

We can represent this behaviour in limit notation as follows:

$\lim_{x\rightarrow\infty}\frac{x^2}{x^2-1}=1$limx→∞x2x2−1​=1 and $\lim_{x\rightarrow-\infty}\frac{x^2}{x^2-1}=1$limx→−∞x2x2−1​=1.

Example 7

Describe the behaviour of $f(x)=\frac{1}{\left(x-1\right)^2}$f(x)=1(x−1)2​ as $x$x approaches $1$1.

From the graph of $f(x)=\frac{1}{\left(x-1\right)^2}$f(x)=1(x−1)2​, we can see a vertical asymptote at $x=1$x=1 and the two branches of the graph approach positive infinity.

We have the limit from left and right in agreement, both approach infinity. So the limit can be written as follows:

 $\lim_{x\rightarrow1}\frac{1}{(x-1)^2}=\infty$limx→11(x−1)2​=∞.

Example 8

Describe the behaviour of $y=\frac{1}{\left(x-1\right)}$y=1(x−1)​ as $x$x approaches $1$1.

For the graph of $y=\frac{1}{x-1}$y=1x−1​ as $x$x approaches $1$1, we see the graph approaching the vertical asymptote in two different directions.

Notice how approaching $x=1$x=1 from the left we have a limit of negative infinity, while approaching $x=1$x=1 from the right we have a limit of positive infinity. As the left and right limits do not agree $\lim_{x\rightarrow1}\frac{1}{x-1}$limx→11x−1​ does not exist.

 

Practice questions

Question 1

Question 2

Question 3

Question 4