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.

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 

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

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.
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 :
$g’(a)$ = $lim_{h\ \rightarrow\ 0}$ $\frac{g(a\ +\ h)\ -\ g(a)}{h}$ = $lim_{h\ \rightarrow\ 0}$ $\frac{f(a\ +\ h,\ b)\ -\ f(a,\ b)}{h}$
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

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