Does the lim of ln(1+x^2y) as (x,y) approach (0,0) exist? If so, is it equal to 0? thanks...?

Answer 1

I have to disagree with Tony:

If you approach (0,0) by setting y = 0 right away, and then decreasing x, you get:
ln(1 + x^(2y)) = ln(1 + exp(2y*ln(x)))
= ln(1 + exp(0*ln(x)))
= ln(1 + exp(0))
= ln(1 + 1)
= ln(2)

On the other hand, if you approach (0,0) by setting x = 0, and then decreasing y, you get:

ln(1 + x^(2y)) = ln(1 + exp(2y*ln(x)))
= ln(1 + exp(2y*ln(0)))
= ln(1 + 0)
= ln(1)
= 0

So since you get two different results by approaching it in two different ways, there is no limit.

Answer 2

The simplest way is to use polar coordinates
ln( 1+x^2*y) = ln ( 1+r^3 cos^2t*sint).
whatever the angle t the limit is 0 if r===>0
if 2y is in the exponent this implies that x has to be >0 and the limit does not exist

Answer 3

This function is continuous at (0,0), so the limit of the function is simply the value of the function at (0,0), namely ln(1 + 0)= 0.