I got $ \lim_{(x,y)\to (1,0)} \ln\left(\frac{1+{y^2}}{{x^2} +xy}\right) $
and I get
- $\lim = 0,$ when $y = 0,$
- $\lim = 0,$ when $y = x,$
but when $y = -x$ I get undefined. So the limit doesn't exist?
I got $ \lim_{(x,y)\to (1,0)} \ln\left(\frac{1+{y^2}}{{x^2} +xy}\right) $
and I get
but when $y = -x$ I get undefined. So the limit doesn't exist?
You need to get straight about what is approaching what:
$ \lim_{(x,y)\to (1,0)} \ln\left(\frac{1+{y^2}}{{x^2} +xy}\right) $
means you need to take the limit of the point $(x, y)\in \mathbb{R}^2$ as $(x, y)$ approaches the point $(1, 0)$, i.e. as both $x\to 1$ AND $y \to 0$.
$ \lim_{(x,y)\to (1,0)} \ln\left(\frac{1+{y^2}}{{x^2} +xy}\right) = \ln\left(\frac{1 + 0^2}{1^2 + 1\cdot 0}\right) = \ln\left(\frac{1}{1}\right) = \ln 1 = 0$.
So no worries.