Bivariate Limit

For bivariate functions, evaluating limits is mostly the same. However, some tools like L’Hôpital’s rule cannot be used.

Definition §

Limit (Bivariate)

Let $f(x,y):D\in {\mathbb{R}}^2\to \mathbb{R}$ be a function of two variables. The limit, $L$ of $f(x,y)$ as $(x,y)$ approaches $(a,b)$ is represented as:

$$\underset{(x,y)\to (a,b)}{\lim }f(x,y)=L$$

if when $(x,y)$ approaches the point $(a,b)$ along every path in $D$, $f(x,y)$ gets close to $L$.

• This is an extension of the limits from different sides, instead of two sides, we must ensure the limit is the same from all paths.

Limit can exist even if function is undefined at $(a,b)$

Just like normal, univariate limits, even if $f(a,b)$ is undefined, the limit $L={\lim }_{(x,y)\to (a,b)}f(x,y)$ may exist.

Evaluating §

Bivariate limits can usually be evaluated the same way normal limits can be. We can use:

• Limit Laws
• Continuity of Limits

Warning L'Hôpital's rule cannot be used to evaluate bivariate limits!

If we get an indeterminate form ($\frac{0}{0}$, $\frac{\infty }{\infty }$, etc.) we can instead change paths.

We can also use the Sandwich Theorem in two dimensions:

Changing Paths §

If we get a limit in indeterminate form (such as $\frac{0}{0}$, $\frac{\infty }{\infty }$, etc. ), we can evaluate the limit by trying different paths to approach it:

• Fix $x$ and vary $y$
• Fix $y$ and vary $x$
• Linearly approach the point $(a,b)$ i.e. $y=\alpha x$ for some $\alpha \ne 0$

If the values are different for different paths, it is sufficient to say the limit does not exist.

Note that this does not prove the limit exists. It can be be used only to disprove the limit.

Examples §

1: Fixing $y$ and $x$

Evaluate the limit:

$$\underset{(x,y)\to (0,0)}{\lim }(\frac{x^2}{x^2+y^2})$$

if it exists.

Solution $0/0$, we can try approaching it from different paths and see if we get a consistent result:

Since this limit is of the indeterminate form

• Fix $y=0$ and vary $x\to 0$:

$$\underset{x\to 0}{\lim }(\frac{x^2}{x^2+0})=\underset{x\to 0}{\lim }(1)=1$$

• Fix $x=0$ and vary $y\to 0$:

$$\underset{y\to 0}{\lim }(\frac{0}{0+y^2})=\underset{y\to 0}{\lim }(0)=0$$

Since the function has different values at different paths, the limit does not exist.

Visuals

2: Linear Paths

Evaluate the limit:

$$\underset{(x,y)\to (0,0)}{\lim }(\frac{xy}{x^2+y^2})$$

if it exists.

Solution $\frac{0}{0}$. We can try approaching the limit through different paths:

Note that the limit is of the indeterminate form

• Fix $y=0$ and $x\to 0$:

$$\underset{x\to 0}{\lim }(\frac{0}{x^2})=0$$

• Fix $x=0$ and $y\to 0$

$$\underset{y\to 0}{\lim }(\frac{0}{y^2})=0$$

• Approach linearly ($y=\alpha x$):

$$\begin{array}{r}\underset{x\to 0}{\lim }(\frac{\alpha x^2}{x^2+{\alpha }^2{x}^2})=\underset{x\to \infty }{\lim }\frac{\cancel{x^2}\cdot \alpha }{\cancel{x^2}(1+{\alpha }^2)}=\underset{x\to 0}{\lim }(\frac{\alpha }{1+{\alpha }^2})=\frac{\alpha }{1+{\alpha }^2}\end{array}$$

Note that this limit changes as $\alpha$ varies, meaning as we approach through different linear paths, the limit changes. Hence, the limit does not exist.

Visuals