Finding the limit $\displaystyle \lim_{(x,y) \to (1,0)} \frac{(x-1)^2 \ln x}{(x-1)^2 + y^2}$ I am asked to find the following limit:
$$ \lim_{(x,y) \to (1,0)} \frac{(x-1)^2 \ln x}{(x-1)^2 + y^2}$$
Even though when I evaluate the limit on the path $y = 0$, I get $0$ (which is the textbook's answer), that still doesn't prove that the limit exists.
$$
\lim_{(x,y) \to (1,0)} \frac{(x-1)^2 \ln x}{(x-1)^2 + y^2} \stackrel{y = 0}{=} \lim_{x \to 1} \ \ln x = 0
$$
I've thought about using $r^2 = x^2 + y^2$ but quickly abandoned the idea (it's not gonna get me anywhere).
Any guidance is highly appreciated.
 A: Let set $u=x-1$ and $v=y$
$$\displaystyle \lim_{(x,y) \to (1,0)} \frac{(x-1)^2 \ln x}{(x-1)^2 + y^2}=\displaystyle \lim_{(u,v) \to (0,0)} \frac{u^2 \ln (u+1)}{u^2 + v^2}=\cos^2\theta\,\ln(r\cos\theta+1)\to 0$$
indeed by squeeze theorem
$$0\le |\cos^2\theta\,\ln(r\cos\theta+1)|\le |\ln(r\cos\theta+1)| \to 0$$
A: Hint:
If $(x,y)\ne (1,0)$, one has
$$0\le\Biggl\lvert\frac{(x-1)^2\ln x}{(x-1)^2+y^2}\Biggr\rvert\le \lvert\ln x\rvert.$$
A: For $x\neq1$, we have 
\begin{eqnarray*}
 &  & \left|\frac{(x-1)^{2}\ln x}{(x-1)^{2}+y^{2}}\right|\\
 & \leq & \left|\frac{(x-1)^{2}\ln x}{(x-1)^{2}}\right|\\
 & = & \left|\ln x\right|.
\end{eqnarray*}
If $x=1$ and $y\neq0$, we have $\left|\frac{(x-1)^{2}\ln x}{(x-1)^{2}+y^{2}}\right|=0=\left|\ln x\right|$.
Hence, for any $(x,y)\in\mathbb{R}^{2}\setminus\{(1,0)\}$, $\left|\frac{(x-1)^{2}\ln x}{(x-1)^{2}+y^{2}}\right|\leq\left|\ln x\right|.$
By Sandwich Rule, $\lim_{(x,y)\rightarrow(1,0)}\frac{(x-1)^{2}\ln x}{(x-1)^{2}+y^{2}}=0$
because $\lim_{(x,y)\rightarrow(1,0)}\left|\ln x\right|=0$.
