limit involving two variables with $\ln(x)$

Question

$$ \lim_{(x,y) \to (1,0)} \frac{(x-1)^2\ln(x)}{(x-1)^2 + y^2}$$

I tried L' Hospitals Rule and got now where. Then I tried using $x = r\cos(\theta)$ and $y = r\sin(\theta)$, but no help. How would I approach this? T

Answer 1

$$\frac{(x-1)^2\ln(x)}{(x-1)^2+y^2} = \frac{\ln(x)}{1+{y^2\over {(x-1)^2}}}.$$ The limit of the numerator is $-\infty$ and the limit of the denominator is $1$ so the limit is $-\infty$.

Answer 2

$$|\frac{(x-1)^2\ln(x)}{(x-1)^2 + y^2}| \leq |\frac{(x-1)^2 \ln (x)}{(x-1)^2}| \stackrel{x \neq 1}{=} |\ln x|$$

But $\ln x \to 0$ as $x \to 1$.

Hence, by squeeze theorem:

$$ \lim_{(x,y) \to (1,0)} \frac{(x-1)^2\ln(x)}{(x-1)^2 + y^2}=0$$