Daniel B. answered 02/24/22
A retired computer professional to teach math, physics
If we follow the given surface to the point (1,0) along two different
curves we get two different limits.
That will prove the claim.
The two different curves are
y = x-1
and
y = √(x-1)
For ease of notation, all limits below are as x->1.
That will also make y->0
When we follow the curve y = x-1
lim (x + (x-1)² - 1) / √[(x-1)² + (x-1)4] =
lim (x-1)x/((x-1)√[1 + (x-1)²]) =
lim x/√[1 + (x-1)²] =
1
When we follow the curve y = √(x-1)
lim (x + (x-1) - 1) / √[(x-1)² + (x-1)²] =
lim 2(x-1) / √[2(x-1)²] =
2/√2