Existence of limit of two variables

Question

According to wolfram alpha math If $f(x,y)=ylog|x| where\ x,y\ are\ real$ then $$\lim_{(x,y)\to(0,0)}f(x,y)= 0 $$ now if we change the given function into polar coordintes then its limit will also be $0$ . i.e, $$\lim_{r\to 0^+}r(sin\theta) log|rcos\theta|=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ .......(1)$$. now the main question is if: $$g(x,y)=|x|^y$$ then its limit at origin is : calculating the limit along the curve $x=e^{(-k/y)}$ as $$\lim_{y\to 0}g(e^{-k/y},y)= \lim_{y\to 0}|e^{(-k/y)}|^y$$ $$\lim_{y\to 0}e^{-k}= 1/e^k$$ means it depends on the value of k. hence limit of $g$ does not exists at origin. BUT THE MAIN PROBLEM IS THAT IF WE CHANGE INTO POLAR COORDINATES THEN CALCULATE ITS LIMIT $$\lim_{r\to 0^+}|rcos\theta|^{rsin\theta}$$ which is $0^0 form$ hence , we write it as $$e^{\lim_{r\to 0^+}r(sin\theta)log|rcos\theta|}$$ from $(1)$ $$\lim_{r\to 0^+}r(sin\theta) log|rcos\theta|=0 $$ therefore $$e^0=1$$ thus by coordinates its limit is unique and finite. which contradicts the limit along curve method. can anyone tell me $$\lim_{r\to 0^+}r(sin\theta) log|rcos\theta|=0$$ it is true or not? if it is false then it contradicts the limit of $f(x,y)=ylog|x|$ can anyone explain it ?

Answer 1

$\lim_{(x,y)\to(0,0)}y\ln(|x|)$ doesn't exist since

$$\lim_{x\to 0}f(x,\frac{k}{\ln(|x|)})=k$$