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Show that
does not exist.
If then Since the answer depends on the limit fails to exist.
To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology.
Let S be a subset of ( [link] ).
Let S be a subset of ( [link] ).
An example of an open set is a disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a disk but not the other half, then the set is neither open nor closed.
Let S be a subset of ( [link] ).
The definition of a limit of a function of two variables requires the disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the is not contained inside the domain. By definition, some of the points of the are inside the domain and some are outside. Therefore, we need only consider points that are inside both the disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point.
Let be a function of two variables, and and suppose is on the boundary of the domain of Then, the limit of as approaches is written
if for any there exists a number such that for any point inside the domain of and within a suitably small distance positive of the value of is no more than away from ( [link] ). Using symbols, we can write: For any there exists a number such that
Prove
The domain of the function is which is a circle of radius centered at the origin, along with its interior as shown in the following graph.
We can use the limit laws, which apply to limits at the boundary of domains as well as interior points:
See the following graph.
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