Multivariable Limit with ln(x)

Question

I need to prove that the following limit is equal to zero:

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

I tried converting to polar coordinates, since this can sort of simplify the denominator, but wound up stuck with $r \cos θ$ inside the logarithmic term. I also tried using the Taylor series for $\ln(x)$, but that did not simplify the expression whatsoever.

How do I go about doing this?

Answer 1

\begin{align}\lim \limits_{(x,y) \to (1,0)} \left|\frac {(x-1)^2\ln(x)}{(x-1)^2 + y^2} \right| &\le \lim \limits_{(x,y) \to (1,0)} |\ln(x)|\\ &=0 \end{align}

Answer 2

Let

• $u=x-1$
• $v=y$

then

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

indeed

$$\frac {u^2 \ln(1+u)}{u^2 + v^2}=\frac{\ln(1+u)}{u}\frac {u^3 }{u^2 + v^2} \to 1\cdot 0=0$$

$$\frac {u^3 }{u^2 + v^2}=r\cdot \cos^3 \theta \to 0$$