Thread 16774913

how does something like this even work? most of the time i can find the trick (it usually amounts to looking at a special line), but i'm never sure of my answer 100% of the time

is there a more surefire but boring way to solve this?

Literally use the definition
lim f(x) = f(c) as x -> c for continuous function

>>16774913 (OP)

my intuition says f1 only

>>16774913 (OP)
bait thread.

>>16774926
If you approach (0,0) on the line x=y, then the function is always 0. If approached on the line x=-y, the function is 1. f1 is not continuous.

f2 is also not continuous, the function is always 1/2 on the line x=y yet it's 0 at the origin.

f3 is continuous, easy to see if you use radial coordinates.

>>16774921
how does the definition work for the first function? it makes perfect sense for the second, but it's indeterminate for the first and there's no l'hopital's rule to be used in multiple variables (i think)

>>16774942
how? what could the bait here possibly be lmao

>>16774944
that's what i did, except for three which is where i got stuck, but i was wondering if there was a technique for this that wasn't just 'find the right line to look at'

>>16774975
Using special lines can't prove that the function is continuous, it can only prove a function is discontinuous. Even if you check every line it's not sufficient, since there might still be a curved path like y=x^2 that gives a different answer.

For f3 just use radial coordinates, x=r cos(theta), y=r sin(theta), then an r factors out and you can easily take the limit.

>>16774913 (OP)
It's going to be quite easy to show to show something is discontinuous. First, check the limit through some direction. Check if it it equals it at that point. If it doesn't you're done. If not, try to find another path that gives another answer. x=0,y=0, y=x^n, x=y^n are common choices.

For the first, y=x instantly tells me f_1 is discontinuous. For the second, see what happens when you approach from the direction y=0 and y=x.

For the third, maybe a few tests might make you think okay this is continuous. You also might think that as the order of the top term is bigger, it goes to 0 faster than the bottom.

As we are approaching the origin, it is convenient to use polar coordinates and the point is, regardless of what function of theta(r) we have, if r approaches 0, does the function approach 0? If so, it is continuous. See where that gets you.

>>16774913 (OP)
1&3
It feels right

>>16774997
Oh not an option. Then only f3 because numerator has higher degrees of zero

>>16774979
the first paragraph is what i've been saying in this thread

my problem is that the second paragraph doesn't always work out either, it's also a special case

>>16775455
Math is full of special cases. Often there isn't (or at least no one knows) a single technique to solve a wide range of problems. You're going to hate higher level maths if you think like that.

If it helps you, usually the problems given to you on exams and exercises are designed in a way such that it's fairly obvious what the tricks are. The problem here has obvious tricks, and for f3 it's intuitive that higher powers on the numerator than on the denominator will force the limit to 0 at the origin. Radial coordinates just makes the argument more concrete.

hey guys how can i remember my multiplication facts im trying to do the number 3

>>16774913 (OP)
I like the polar coordinate test

>>16776872
this is equivalent to just inspecting powers

>>16776872
I just realized I wrote "not equals" twice per each piecewise function. I'm quite a goofy goober.