Q. 15 - Calculus

Question

Provide a definition for $\underset{(x,y)\to (a,b)}{\lim }f(x,y)=\infty$. Model your definition on Definitions 1.9 and 12.15. 

Step-by-Step Solution

Answer

For all $M>0$,there exist a $\delta >0$ such that if $0<\sqrt{{(x-a)}^2+{(y-b)}^2}<\delta$, then $f\left(x\right)\in (M,\infty )$.

Step 1: Defining the limit

The goal is to provide a definition for $\underset{(x,y)\to (a,b)}{\lim }f(x,y)=\infty$.

The infinite limit at a point, represented as $\underset{\left(x\right)\to \left(a\right)}{\lim }f\left(x\right)=\infty$, denotes the existence of an infinite limit for every $M>0$, such that if $x\in (a-\delta ,a)\cup (a,a+\delta )$, then $f\left(x\right)\in (M,\infty )$

The limit of a two-variable function f is L, expressed as $\underset{\left(x\right)\to \left(a\right)}{\lim }f\left(x\right)=L$ if, for every $\epsilon >0$, there is $\delta >0$ such that $\left|f\left(x\right)-L\right|<\epsilon$ whenever $0<\left|\left|x-a\right|\right|<\delta$.

Step 2: Evaluating the limit

To define the infinite limit of a function in two variables, combine the two definitions.

The infinite limit of a function in two variables at a point, represented as $\underset{(x,y)\to (a,b)}{\lim }f(x,y)=\infty$, indicates that for any $M>0$, there exists $\delta >0$ such that if is $0<\sqrt{{(x-a)}^2+{(y-b)}^2}<\delta$, then $f\left(x\right)\in (M,\infty )$.