The definition of a limit

Functions: Limits and asymptotes

The definition of a limit

We have seen that quotient functions can be undefined for one or more values. By using limits, we can still investigate what happens at these values.

Limit for a value

The limit for a point $x=\green a$ of a function $\blue{f(x)}$ is the value the function approximates as we get closer to $\green a$ . We don't have to reach the value, but as long as we get closer to $\green a$ , we get a better approximation of the value of the limit.

We denote the limit of $x$ to a point $\green a$ of a function $\blue{f(x)}$ as $\lim_{x \to \green a}\blue{f(x)}$ .

It might matter for the value of a limit whether we approximate the $x$ -value $\green a$ from above or from below. That is why we distinguish the right limit and the left limit.

The right limit is the limit if we approximate $x=\green a$ from above. This means that we get closer to point $\green a$ from a value greater than $\green a$ . We denote this as $\lim_{x \downarrow \green a}\blue{f(x)}$ .

The left limit is the limit if we approximate $x=\green a$ from below. This means that we get closer to point $\green a$ from a value less than $\green a$ . We denote this as $\lim_{x \uparrow \green a}\blue{f(x)}$ .

If both the left limit and the right limit at a point $x= \green a$ are equal, we call this the limit at $x=\green a$ .

Examples

$\begin{array}{rcl} \lim\limits_{x \to \green2}\blue{\frac{x^2+x-6}{x-2}}&=& \lim\limits_{x \to \green2} {x+3}\\&=&5 \\ \\ \lim\limits_{x \downarrow \green1}\blue{\frac{1}{x-1}}&=&\infty \\ \\ \lim\limits_{x \uparrow \green1}\blue{\frac{1}{x-1}}&=&-\infty \\ \\ \lim\limits_{x \to \green{-5}}\blue{\frac{x^3-5x^2+5x-25}{x+5}}&=& \lim\limits_{x \to \green{-5}} {x^2+5}\\&=&30 \\ \\ \lim\limits_{x \downarrow \green2}\blue{\frac{1}{x^2-4}}&=&\infty \\ \\ \lim\limits_{x \uparrow \green2}\blue{\frac{1}{x^2-4}}&=&-\infty \end{array}$

Calculating limits

The limit at a point $x=\green a$ is easy to find if the function is defined at this point. In that case, the limit is equal to the function value in that point.

If the function is not defined at $x=\green a$ , because the denominator would equal 0 at $x=\green a$ , it is a bit more complicated.

If both the numerator and the denominator are equal to $0$ in $x=\green a$ , it is not hard to calculate the limit. In this case, we simply divide both the numerator and the denominator by $x-\green a$ . This way, the function will be simplified to an expression in which $x=\green a$ is defined. In which case the limit equals the function value at that point.

If, however, only the denominator equals $0$ in $x=\green a$ , then the limit will probably be $\pm\infty$ . This will be explained in tab "infinity".

It might happen that the function is not defined for other reasons. These cases are beyond the scope of this course.

Example

$\begin{array}{rcl}\lim\limits_{x \to \green{-2}}\blue{\frac{x^2-2x+1}{x-4}}&=&\frac{(\green{-2})^2-2\cdot \green{-2}+1}{\green{-2}-4}\\&=&-\frac{3}{2} \\ \\ \lim\limits_{x \to \green4}\blue{\frac{x^2-6x+8}{x-4}}&=&\lim\limits_{x \to \green4}\blue{\frac{(x-2) \cdot (x-4)}{x-4}} \\ &=&\lim\limits_{x \to \green4} {x-2}\\&=&2 \\ \\ \lim\limits_{x \to \green{-3}}\blue{\frac{x^2+5x+6}{x+3}}&=&\lim\limits_{x \to \green{-3}}\blue{\frac{(x+2) \cdot (x+3)}{x+3}} \\ &=&\lim\limits_{x \to \green{-3}} {x+2}\\&=&-1 \end{array}$

Infinity

When close to $x=\green a$ only the denominator tends to $0$ and not the numerator, or the denominator decreases faster than the numerator, then the limit tends to infinity or minus infinity.

Example

$\begin{array}{rcl} \lim\limits_{x \downarrow \green3}\blue{\frac{x}{x^2-9}}&=& \infty \end{array}$

Numerically

The limit at a point $x=\green a$ of a function $\blue{f(x)}$ is the value that the function approximates when we get closer to $\green a$ . We can demonstrate this behavior numerically.

Consider the function $\blue{f(x)}=\blue{\frac{x^2-4}{x-2}}$ . This function is not defined at $x=\green{2}$ , but $\lim_{x \to \green 2}\blue{\frac{x^2-4}{x-2}}=\lim_{x \to \green 2}\blue{ \left( x+2 \right)} = 4$ . This implies that when we approach $x=\green 2$ the value of the function tends to $4$ . We can see this in the next table.

$x$	$\blue{f(x)}$
$1$	$3$
$1.5$	$3.5$
$1.9$	$3.899999\ldots$
$1.99$	$3.989999\ldots$
$1.999$	$3.998999\ldots$
$1.9999$	$3.999900\ldots$
$1.99999$	$3.999989\ldots$
$1.999999$	$3.999998\ldots$

The table shows that as we approach $x=\green 2$ the value of $\blue{f(x)}$ gets closer to $4$ .

In different literature, you may encounter other notations for the right and left limits. The right limit, which we denote by $\lim_{x\downarrow a}f(x)$ , can also be written as

$\lim_{x\,\searrow\, a}f(x) \quad\text{or}\quad \lim_{x\to a^+}f(x).$ The left limit, which we denote by $\lim_{x\uparrow a}f(x)$ , can also be written as

$\lim_{x\,\nearrow\, a}f(x) \quad\text{or}\quad \lim_{x\to a^-}f(x).$

We can also investgate what happens if $x$ gets really big or small. This is the limit to infinity or to minus infinity.

Limit to infinity

We can consider the limit of $x$ to $\green{\infty}$ or to $\green{-\infty}$ of $\blue{f(x)}$ . In that case, we want to find out the value of $\blue{f(x)}$ when $x$ ever increases or decreases.

We denote this as $\lim_{x \to \green{\infty}}\blue{f(x)}$ for the limit of $x$ to $\green{\infty}$ or as $\lim_{x \to \green{-\infty}}\blue{f(x)}$ for the limit of $x$ to $\green{-\infty}$ .

Examples

$\begin{array}{rcl}\lim\limits_{x \to \green{\infty}} \blue{\frac{1}{x+1}}&=&0 \\ \\ \lim\limits_{x \to \green{-\infty}}\blue{\frac{x}{x^2+1}}&=&0 \end{array}$

Calculating

	Step-by-step	Example
	We calculate the limit of a quotient function $\blue{f(x)}=\blue{\frac{p(x)}{q(x)}}$ to $\green{ \pm\infty}$ .	$\lim\limits_{x \to \green{-\infty}}\blue{\frac{-3x^2+1}{x+4}}$
Step 1	Put the largest power of $x$ in the denominator in front of the brackets in both the numerator and the denominator	$\lim\limits_{x \to \green{-\infty}}\blue{\frac{x\cdot\left(-3x+\frac{1}{x}\right)}{x\cdot\left(1+\frac{4}{x}\right)}}$
Step 2	Divide both the numerator and the denominator by the factor in front of the brackets.	$\lim\limits_{x \to \green{-\infty}}\blue{\frac{-3x+\frac{1}{x}}{1+\frac{4}{x}}}$
Step 3	Check if there are terms $\frac{a}{x^p}$ with $a$ a number and $p\geq 1$ in either the numerator or the denominator. These tend to $0$ when $x$ goes to $\pm \infty$ .	$\lim\limits_{x \to \green{-\infty}}{\frac{-3x}{1}}$
Step 4	Determine the limit by reasoning what happens to the terms that you have left when $x$ tends to $\pm \infty$ .	$\infty$

Given the function $f(x)=\frac{x^2+3\cdot x-18}{x^2+2\cdot x-24}$ . Calculate $\lim_{x\downarrow \, -6 } f(x)$ and simplify. If the limit doesn't exist, write $none$ .

$\lim_{x\downarrow \, -6 } \frac{x^2+3\cdot x-18}{x^2+2\cdot x-24}= {{9}\over{10}}$

$\begin{array}{rcl} \displaystyle \lim_{x\downarrow \, -6 } f(x) &=& \displaystyle\lim_{x\downarrow \, -6 } \frac{x^2+3\cdot x-18}{x^2+2\cdot x-24} \\ && \qquad \blue{\text{function } f(x) \text{ substituted}}\\ &=& \displaystyle\lim_{x\downarrow \, -6 } \frac{(x+6)\cdot(x-3)}{(x+6)\cdot(x-4)}\\ && \qquad \blue{f(x) \text{ factored}}\\ &=& \displaystyle\lim_{x\downarrow \, -6 } \frac{(x-3)}{(x-4)}\\ && \qquad \blue{\text{divide numerator and denominator by } x+6 }\\ &=& \displaystyle \frac{(-6-3)}{(-6-4)}\\ && \qquad \blue{\text{the quotient function } \frac{(x-3)}{(x-4)} \text{ is defined in } x=-6 \text{, so this point is substituted}}\\ &=& \displaystyle {{9}\over{10}}\\ && \qquad \blue{\text{calculated}}\\ \end{array}$

New example