Thanks for the answer

@yswong welcome

For part (a), the method of y=mx can only prove discounity, it cannot be used to prove that the limit exists?

@yswong yes. more precisely, you cannot use y=mx to disprove continuity, since it has limit 0. to disprove continuity, you have to take y=mx^2 in this case. by the way, you can use x=rcos(t) and y=rsin(t) to prove continuity

But again part(a) mentions only on a straight line, does the method of y=mx^2 or polar coords works on a straight line?

i dont understand what you are asking.

oh hi

I think i will just simply use polar coords to solve this question for part (a)

Thanks for ur time

i doubt whether you can use polar coordinate to do part a

usually we said y=mx cannot prove continuity because it only shows the limit exist along straight lines

and polar coords is usually used to prove continuity in all cases

In this cases, the limit exist along straight lines, but not in all cases such as parabola(y=mx^2)

it is hard to use polar coords

this question is asking is that even though the limits clearly does not exists, we have to prove why it does seem to exists at 0 when we only consider the approach path of the straight line.

Okay thanks i understood ur explanation already