Introduction to limits
What is a limit?
All of calculus starts with one idea: the limit. A limit tells us what value a function approaches as $x$ approaches a number. (Approaches means gets closer and closer to.) What matters is where the function is heading. What happens exactly at that number does not matter.
Look at the function
$f(x)=\dfrac{x^2-1}{x-1}$.
At $x=1$, the numerator and the denominator are both $0$. So $f(1)=\dfrac{0}{0}$, which is undefined. But we can still ask: what happens to $f(x)$ when $x$ is close to $1$?
The table shows $f(x)$ for values of $x$ close to $1$. Some are a little less than $1$ (the left side). Some are a little more than $1$ (the right side).
| $x$ | $f(x)$ |
|---|---|
| 0.9 | 1.9 |
| 0.99 | 1.99 |
| 0.999 | 1.999 |
| 1.001 | 2.001 |
| 1.01 | 2.01 |
| 1.1 | 2.1 |
As $x$ approaches $1$ from both sides, $f(x)$ approaches $2$. So we say: the limit of $f(x)$ as $x$ approaches $1$ is $2$.
The graph shows why. When $x$ is not $1$, we can factor the numerator and cancel the common factor:
$f(x)=\dfrac{(x-1)(x+1)}{x-1}=x+1$, for $x\neq 1$.
So the graph of $f$ is the line $y=x+1$ with one point missing. The missing point is at $(1,2)$. We show it with an open circle, called a hole.
Limit notation
We write
$\displaystyle\lim_{x\to a}f(x)=L$
We read this as "the limit of $f(x)$, as $x$ approaches $a$, is $L$."
It means: when $x$ is close to $a$, $f(x)$ is close to $L$. We can make $f(x)$ as close to $L$ as we want by taking $x$ close enough to $a$. The values of $x$ can come from either side of $a$, but $x$ is never equal to $a$.
We can also write the same thing as
$f(x)\to L$ as $x\to a$.
Read the arrow as "approaches." For the function above,
$\displaystyle\lim_{x\to 1}\dfrac{x^2-1}{x-1}=2$.
The value at the point does not matter
To find $\displaystyle\lim_{x\to a}f(x)$, we never use $x=a$ itself. Sometimes the function is undefined at $a$, like the example above. Sometimes it has a value at $a$ that is different from the limit.
For example, look at this function $g$:
$g(x)=\begin{cases}x+1, & x\neq 1\\3, & x=1\end{cases}$
This means $g(x)=x+1$ for every $x$ except $1$, and $g(1)=3$.
Near $x=1$, the graph of $g$ is the same line as before. So $\displaystyle\lim_{x\to 1}g(x)=2$, even though $g(1)=3$.
In short: the limit tells us where $f(x)$ is heading as $x$ approaches $a$. The value $f(a)$ tells us where the function actually is at $a$. These two can be the same or different. Also, one of them may not exist.
Estimating a limit from a table
To estimate a limit with a table, pick values of $x$ closer and closer to $a$, from both sides. Find $f(x)$ for each one. Then see what number the values of $f(x)$ approach.
Example 1: Estimate $\displaystyle\lim_{x\to 2}\dfrac{x^3-8}{x-2}$.
Solution:
The function is undefined at $x=2$. So we find its values at numbers close to $2$:
| $x$ | $\dfrac{x^3-8}{x-2}$ |
|---|---|
| 1.9 | 11.41 |
| 1.99 | 11.9401 |
| 1.999 | 11.994001 |
| 2.001 | 12.006001 |
| 2.01 | 12.0601 |
| 2.1 | 12.61 |
From both sides, the values approach $12$. So,
$\displaystyle\lim_{x\to 2}\dfrac{x^3-8}{x-2}=12$.
Example 2: Estimate $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}$, where $x$ is in radians.
Solution:
Set your calculator to radian mode.
The function gives the same value at $x$ and at $-x$, because $\dfrac{\sin(-x)}{-x}=\dfrac{\sin x}{x}$. So one table covers both sides:
| $x$ | $\dfrac{\sin x}{x}$ |
|---|---|
| ±0.1 | 0.998334 |
| ±0.01 | 0.999983 |
| ±0.001 | 0.9999998 |
The values approach $1$. So,
$\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$.
This limit is very important in calculus. We will prove it later, using the Squeeze Theorem.
Calculator tip: In calculus, we measure angles in radians, unless a problem says otherwise. So set your calculator to radian mode before you work with trigonometric functions (sin, cos, tan). In degree mode, you will get different numbers. For example, in Example 2 the values would approach $\dfrac{\pi}{180}\approx 0.01745$, not $1$. And $\sin\left(\dfrac{\pi}{0.1}\right)$ in the next paragraph would give $\sin(31.4159^\circ)\approx 0.5212$, not $0$.
A table can mislead. A table shows only a few values. The function can do something different between those values.
For example, take $f(x)=\sin\left(\dfrac{\pi}{x}\right)$, with the calculator in radian mode. At $x=0.1$, $0.01$, and $0.001$, the values are all exactly $0$. So it looks like the limit is $0$ as $x\to 0$. But $f(0.4)=1$ and $f(\tfrac{2}{9})=1$. In fact, the function keeps oscillating (going up and down) between $-1$ and $1$, no matter how close $x$ gets to $0$. So the limit does not exist.
Use a table to make a first guess. Then check it with a graph or with algebra.
Estimating a limit from a graph
To read the limit at $x=a$ from a graph, that is, to find $\displaystyle\lim_{x\to a}f(x)$:
- Find $x=a$ on the $x$-axis.
- Move along the graph toward $x=a$ from the left. Note the height ($y$ value) that the graph approaches.
- Move along the graph toward $x=a$ from the right. Note the height that the graph approaches.
- If both sides approach the same height $L$, then the limit is $L$: $\displaystyle\lim_{x\to a}f(x)=L$.
Only the graph near $x=a$ matters. An open circle or a solid point right at $x=a$ does not change the limit.
Example 3: Use the graph of $h$ to find $\displaystyle\lim_{x\to -2}h(x)$, $\displaystyle\lim_{x\to 2}h(x)$, and $h(2)$.
Solution:
As $x$ approaches $-2$ from both sides, the graph approaches the height $1$. So $\displaystyle\lim_{x\to -2}h(x)=1$. Here the limit is the same as the function value: $h(-2)=1$.
As $x$ approaches $2$ from both sides, the graph approaches the open circle at height $1$. So $\displaystyle\lim_{x\to 2}h(x)=1$.
The solid point at $x=2$ shows that $h(2)=3$. So here the limit and the function value are different.
When a limit does not exist
A limit exists only when $f(x)$ approaches one single number as $x$ approaches $a$ from both sides. Here are three common ways this can fail:
- The graph jumps. From the left, $f(x)$ approaches one number. From the right, it approaches a different number (figure a). You will learn more about this in one-sided limits.
- The function increases without bound (gets larger and larger), as $\dfrac{1}{x^2}$ does near $x=0$ (figure b). You will learn more about this in infinite limits.
- The function keeps oscillating and never settles at one value, as $\sin\left(\dfrac{\pi}{x}\right)$ does near $x=0$ (figure c).
Optional
The precise definition of a limit
The words "close to" and "approaches" can be made exact.
Let $f$ be defined on an open interval that contains $a$. ($f$ does not need to be defined at $a$ itself.)
An open interval is all the numbers between two endpoints, without the endpoints. For example, the open interval $(2,5)$ is all $x$ with $2<x<5$. So "$f$ is defined on an open interval that contains $a$" means: $f$ is defined at every $x$ a little to the left of $a$ and a little to the right of $a$. This makes sure that $x$ can approach $a$ from both sides.
Then
$\displaystyle\lim_{x\to a}f(x)=L$
means: for every number $\varepsilon>0$, there is a number $\delta>0$ such that
if $0<|x-a|<\delta$, then $|f(x)-L|<\varepsilon$.
Here $|x-a|$ is the distance between $x$ and $a$, and $|f(x)-L|$ is the distance between $f(x)$ and $L$.
In plain words: pick any small distance $\varepsilon$ around $L$. Then you can always find a distance $\delta$ around $a$ that works like this: every $x$ within $\delta$ of $a$ (except $a$ itself) gives an $f(x)$ within $\varepsilon$ of $L$.
Example: Show that $\displaystyle\lim_{x\to 3}(2x-1)=5$.
We want $|(2x-1)-5|<\varepsilon$. First, simplify the left side:
$|(2x-1)-5|=|2x-6|=2|x-3|$.
So we need $2|x-3|<\varepsilon$. Divide both sides by $2$:
$|x-3|<\dfrac{\varepsilon}{2}$.
So choose $\delta=\dfrac{\varepsilon}{2}$. Then whenever $0<|x-3|<\delta$, we get $|f(x)-5|<\varepsilon$. This works for every $\varepsilon>0$, so the limit is proved.