- #1

demonelite123

- 219

- 0

(x,y)->(0,0)

i just came up with this limit to test my understanding. so approaching from y = mx, y = mx^2, and y = mx^3 all seemed to point out that the limit is 0. i tried approaching from the x and y axes and i got for approaching along the x axis: lim (as (x,y)->(0,0)) of 0 / x. would this be another instance of the limit being 0 or would it be an indeterminate form? because x is approaching 0 but it does not equal 0 so x would get smaller and smaller yet still be a finite number greater than 0 so the quotient would be 0. is this reasoning correct?

i have also tried polar coordinates and came up with: lim (as r->0) of (r^2 cos(x)sin(x))/(cos(x) + r sin^2 (x)). so when cos and sin are nonzero the limit becomes 0, no problem. when x = 0 the limit is also 0. but when x = pi/2 then i get lim (as r->0) of 0 / r which brings me back to the question i asked earlier about 0 / x. is this 0 or would it be indeterminate?

i know i can't say for sure that this limit exists but i'm starting to suspect it does if the 0 / x and 0 / r quotients indeed are also 0 when x and r approach 0 respectively.