Limits

$\lim_{(x,y)\to{(a,b)}} {f(x)}=L$ iff: $$

$$\forall{\epsilon>0}\; \exists{\delta>0}\; \forall{(x,y)\in D}\; \bigg( 0<\sqrt{(x-a)^2 + (y-b)^2}<\delta\implies |f(x,y)-L|<\epsilon \bigg)$$

In the above definition statement, circular $\delta-$disk was used; square neighbourhood can also be used. The target point can be approached in any directions. $$

Multivariable limit properties are analogous to the single variable limits.

Uniqueness of limit

Let $f$ be a real-valued function defined on $D\subseteq \mathbb{R}^2$. Let $(a,b)\in \overline{D}$.

If $\lim_{(x,y)\to {(a,b)}} f$ exists, then it is unique. $$

Proof Hint

• Consider 2 limits $l,m$ for $f$ at $(a,b)$
• Consider $\delta_1, \delta_2 > 0$ for both limitting values for a chosen $\epsilon \gt 0$
• Reduce to $\forall \epsilon \gt 0, |l-m| < \epsilon$

Non existence of limit

Suppose as $(x,y) \to (a,b)$, $f \to L_1$ along a path $C_1$ and $f\to L_2$ along a different path $C_2$.

If $L_1 \neq L_2$ then the $\lim_{(x,y)\to (a,b)} f$ doesn’t exist.

Iterated limits

Aka. repeated limits. If $f$ is defined in a neighborhood of a point $(a,b)$ in $\mathbb{R}^2$ and $\lim_{x\to a} f(x,y)$ exists, which is a function of $y$ only, then the limit of this function as $y\to b$ can be written as:

$$\lim_{y\to{b}} \; \lim_{x\to{a}} {f(x,y)}$$

Similarily, another limit exists.

$$\lim_{x\to{a}} \; \lim_{y\to{b}} {f(x,y)}$$

Note:

• The two repeated limits may or may not exist independently
• The two repeated limits, when they exist, may or may not be equal.
• Existence of the 2-variable limit $\centernot\implies$ existence of either of the two repeated limits
• Existence of the repeated limits $\centernot\implies$ existence of 2-variable limit
• If the repeated limit exists and they are not equal, then the 2-variable limit cannot exist.
• If a repeated limit exists, along with the 2-variable limit, then these two would be equal