Limit of a function is the value to which the function is approaching, with the input of the function approaches to some value.
It is written as

$Lim_{(x\ \rightarrow\ c)}\ f(x)\ =\ L$

Here, $f\ (x)$ is a function whose limit is calculated as $x$ approaches to $c$ and the limit is found to be $L$. In this article, we are going to discuss about the limit of a function of two variables. We shall also see its definition and related formulas. Let us go ahead and learn about this concept in detail.

## Definition

Now, let $f$ be a function in two variables $x$ and $y$ such that $f\ (x, y)$. Also, let $f$ be defined in a circular region around the point $(x_{0},\ y_{0})$. Then, we say that limit of $f\ (x, y)$ is approaching to $L$ if and only if for every epsilon greater than zero, there is existing a delta greater than zero such that  the function satisfies the following condition:

$|\ f\ (x, y)\ –\ L\ |\ <\ \varepsilon$

With the condition that the distance between $(x_{0},\ y_{0})$ and $(x,\ y)$ is also satisfying

$0\ < \sqrt{(x\ –\ x_{0})^{2}\ +\ (y\ –\ y_{0})^{2}}\ <\ \delta$

The commonly used notation in such case for limits if it exists is

$Lim_{(x,\ y) \rightarrow (x_{0},\ y_{0})}\ f\ (x,\ y) = L$

## Formulas

If we have,

$lim_{(x,\ y)\ \rightarrow\ (x_{0}, y_{0})}$ $f\ (x,\ y)\ =\ K$,

$lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}$ $g\ (x,\ y)\ =\ L$

and ‘$c$’ is any real number, then the following results are true.
1) $Lim_{(x,\ y)\ \rightarrow\ (x_{0}, y_{0})}$ $[c\ \times\ f\ (x,\ y)]$ = $c\ Lim_{(x,\ y)\ \rightarrow\ (x_{0}, y_{0})}$ $f\ (x,\ y)$

This is called linearity 1.
2) $Lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}$ $[(f\ +\ g)\ (x,\ y)]$ = $Lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}$ $f\ (x,\ y)$ + $Lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}$ $g\ (x,\ y)$

This is linearity 2.
3) $Lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}$ $[(f\ g)\ (x,\ y)]$ = $[Lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}$ $f\ (x,\ y)]$ $[Lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}$ $g\ (x,\ y)$

This is product of functions property.
4) $Lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}$ $[\frac{f\ (x,\ y)}{g\ (x,\ y)}]$ = $[\frac{Lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}\ f(x,\ y)}{[Lim_{(x,\ y)\ \rightarrow\ (x_{0},\ y_{0})}\ g(x,\ y)}]$

This is quotient of functions property, but in this case the only condition is L must not be equal to zero.
The results of linearity and the product of functions listed above can be generalized for any finite number of functions as below:

1) The limit of the sum of $‘n’$ functions is the same as the sum of the limits of $‘n’$ functions.

2) The limit of the product of $‘n’$ functions is the same as the product of the limits of $‘n’$ functions.

For these results to be valid, it is important that the limit of all the functions must exist individually.

## Limits & Derivatives

The limits and derivatives are interconnected with each other. It is because the definition of derivative is given in terms of limits of a function. Let us suppose a function of two variables $f(x,\ y)$ and $(a,\ b)$ be a point in its domain.

While talking about the direction of $X,\ y$ is held and fixed at $y\ =\ b$. Therefore, the function $f(x,\ b)$ is the required function in variable $x$, we may write $f(x,\ b)\ =\ g(x)$, where $g(x)$ is a differentiable function at $x\ =\ a$. In such case, we may define derivative in the following way :
$g’(a)$ = $lim_{h\ \rightarrow\ 0}$ $\frac{g(a\ +\ h)\ -\ g(a)}{h}$ = $lim_{h\ \rightarrow\ 0}$ $\frac{f(a\ +\ h,\ b)\ -\ f(a,\ b)}{h}$

## Limits of Multivariable Functions

Suppose we have a three variable function $f\ (x,\ y,\ z)$ which is centered on the point $(x_{0},\ y_{0},\ z_{0})$. Then we say that the limit of the function $f\ (x,\ y,\ z)$ is $L$ as the point $(x,\ y,\ z)$ approaches to the point $(x_{0},\ y_{0},\ z_{0})$, only if for every epsilon greater than zero, there exists a delta greater than zero in such a way that the function satisfies
$|\ f\ (x,\ y,\ z)\ –\ L\ |\ <\varepsilon$

With the condition that the distance between the points $(x,\ y,\ z)$ and $(x_{0},\ y_{0},\ z_{0})$ satisfies the condition

$0\ <\ \sqrt{[(x\ –\ x_{0})^{2}\ +\ (y\ –\ y_{0})^{2}\ +\ (z\ –\ z_{0})^{2}]}$ $< \delta$

Also, we can then write that

$Lim_{(x,\ y,\ z)\ \rightarrow\ (x_{0},\ y_{0},\ z_{0})}$ $f\ (x,\ y,\ z)\ =\ L$

or

$Lim_{(x,\ y,\ z)\ \rightarrow\ (x_{0},\ y_{0},\ z_{0})}$ $|\ f\ (x,\ y,\ z)\ –\ L|\ =\ 0$

## Example

Let us see an example on the described concepts to understand them more.
Example 1:

Check if the limit of the given function exists or not. If the limit exists, find it.

$Lim_{(y,\ z)\ \rightarrow\ (0,\ 0)}$ $[\frac{(y^{3}\ z)}{(y^{6}\ + z^{2})}]$

Solution:

$Lim_{(y,\ z)\ \rightarrow\ (0,\ 0)}$ $[\frac{(y^{3}\ z)}{(y^{6}\ + z^{2})}]$

At the point $(0,\ 0)$ we have continuity problems with the given function. Hence, we will find different paths giving different values of limits if possible.

Consider $z\ =\ y$. Then we have,

$Lim_{(y,\ z)\ \rightarrow\ (0,\ 0)}$ $[\frac{(y^{3}\ z)}{(y^{6}\ +\ z^{2})}]$

= $Lim_{(y,\ y)\ \rightarrow\ (0,\ 0)}$ $[\frac{(y^{3}\ y)}{(y^{6}\ +\ y^{2})}]$

= $Lim_{(y,\ y)\ \rightarrow\ (0,\ 0)}$ $[\frac{(y^{4})}{(y^{6}\ +\ y^{2})}]$

= $Lim_{(y,\ y)\ \rightarrow\ (0,\ 0)}$ $[\frac{(y^{2})}{(y^{4}\ +\ 1)}]$

= $0$

Again, consider $z\ =\ y^{3}$

$Lim_{(y,\ z)\ \rightarrow\ (0,\ 0)}$ $[\frac{(y^{3}\ z)}{(y^{6}\ +\ z^{2})}]$

= $Lim_{(y,\ y^{3})\ \rightarrow\ (0,\ 0)}$ $[\frac{(y^{3}\ y^{3})}{(y^{6}\ +\ (y^{3})^{2})}]$

= $Lim_{(y,\ y^{3})\ \rightarrow\ (0,\ 0)}$ $[\frac{(y^{6})}{(y^{6}\ +\ y^{6})}]$

= $Lim_{(y,\ y^{3})\ \rightarrow\ (0,\ 0)}$ $[\frac{1}{2}]$

= $\frac{1}{2}$

Since we have different paths with different limits for the given function, hence the limit of the given function does not exist.