To understand the concept of limits for multivariable functions, let us have a look at the limits of functions of one variable.
An open interval about "c" is given by $(d-\delta,c+\delta)$, where $\delta$ is some real number.
An open interval about $5$ for $\delta = 2$ is $(3,7)$.
A function $f(x)$ is defined on an open interval about $c$, except possibly at $c$. If the value of $f(x)$ is sufficiently close to $L$ for all $x$ sufficiently close to $c$, then we say that $f$ approaches limit $L$ as $x$ approaches $c$. It is given as follows:
$$\lim_{x \rightarrow c} f( x) =L$$
There are two ways to approach $c$,
- from its left
- from its right
Let us look at two functions and observe the value of $f(x)$ as $x$ becomes sufficiently close to $3$.
Let us look at the value of $f(x)$ when $x$ is sufficiently close to $c$, where $c=3$.
| x |
2.91 |
2.92 |
2.93 |
2.94 |
2.95 |
2.96 |
2.97 |
2.98 |
2.99 |
| f(x) |
1.0081 |
1.0064 |
1.0049 |
1.0036 |
1.0025 |
1.0016 |
1.0009 |
1.0004 |
1.0001 |
| x |
3.09 |
3.08 |
3.07 |
3.06 |
3.05 |
3.04 |
3.03 |
3.02 |
3.01 |
| f(x) |
1.0081 |
1.0064 |
1.0049 |
1.0036 |
1.0025 |
1.0016 |
1.0009 |
1.0004 |
1.0001 |
From both the sides, we can see that the value of $f(x)$ is sufficiently close to $1$ as $x$ becomes sufficiently to $3$.
| x |
2.91 |
2.92 |
2.93 |
2.94 |
2.95 |
2.96 |
2.97 |
2.98 |
2.99 |
| g(x) |
-1 |
-1 |
-1 |
-1 |
-1 |
-1 |
-1 |
-1 |
-1 |
| x |
3.09 |
3.08 |
3.07 |
3.06 |
3.05 |
3.04 |
3.03 |
3.02 |
3.01 |
| g(x) |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
From the left, we can see that $g(x)$ is $-1$ when $x$ is sufficiently close to $3$.
From the right, we can see that $g(x)$ is $1$ when $x$ is sufficiently close to $3$.
The function $f(x)=(x-3)^2+1$ satisfied the informal description of limit. The value of $f(x)$ was sufficiently close to $1$ for all $x$ sufficiently close to $3$. Therefore,
$$\lim_{x\rightarrow 3} f(x) = \lim_{x\rightarrow 3} (x-3)^2+1=1$$
The function $g(x)=\frac{|x-3|}{x-3}$ did not satisfy the informal description of limit. When $x$ was sufficient close to $3$, the value of $g(x)$ was not sufficiently close to any "single" $L\in ℝ$. Therefore, $\lim_{x\rightarrow 3}g(x)$ does not exist.
To understand this in a better way, let us look at the graph of these two functions.
| $f(x)=(x-3)^2+1$ | $g(x)=\frac{|x-3|}{x-3}$ |
|
|
On taking a closer look at $g(x)=\frac{|x-3|}{x-3}$, we can see that the following are true:
- From the left, $g(x)$ is $-1$ for all $x$ sufficiently close to $3$.
- From the right, $g(x)$ is $1$ for all $x$ sufficiently close to $3$.
These two observations will help us to informally describe "Left Limit" and "Right Limit".
A function $f(x)$ is defined on an open interval about $c$, except possibly at $c$. If the value of $f(x)$ is sufficiently close to $L$ for all $x<c$ sufficiently close to $c$, then we say that the left limit of $f$ exists at $c$ and is equal to $L$. It is given as:
$$\lim_{x\rightarrow c^-}f(x)=L$$
A function $f(x)$ is defined on an open interval about $c$, except possibly at $c$. If the value of $f(x)$ is sufficiently close to $L$ for all $x>c$ sufficiently close to $c$, then we say that the right limit of $f$ exists at $c$ and is equal to $L$. It is given as:
$$\lim_{x\rightarrow c^+}f(x)=L$$
A function $f(x)$ is defined on an open interval about $c$, except possibly at $c$. We say that the limit of $f(x)$ at $c$ exists and is equal to $L$ if $:$
- Left hand limit of $f$ exists at $c$ and is equal to $L$
- Right hand limit of $f$ exists at $c$ and is equal to $L$
As mentioned in the introductory page, these notes are aimed at providing an intuition about the concepts. To formally understand what "sufficiently close " means, kindly have a look at the EPSILON-DELTA definition of limit.