(i) Yes. Simplify
.

Now compute the limit by converting to polar coordinates.

This tells us

so we can define
to make the function continuous at the origin.
Alternatively, we have

and

Now,


so by the squeeze theorem,

and
approaches 1 as we approach the origin.
(ii) No. Expand the fraction.

and
are undefined, so there is no way to make
continuous at (0, 0).
(iii) No. Similarly,

is undefined when
.
Answer: 4
Step-by-step explanation:
(Can someone please answer my math question)
.........................It is 40.