Evaluating $\lim\limits_{(x,y) \to (0,0)}x\ln\sqrt{\smash[b]{x^2+y^2}}$

Question

Evaluate $$\lim_{(x,y) \to (0,0)}x\ln\sqrt{\smash[b]{x^2+y^2}}.$$

Attempt. It is indefinite form is $0(-\infty)$. For curves $x=0,~y=0,~y=x$ the limit equals $0$, so the limit could be $0$, if it exists. But I find it hard to estimate $|x\ln\sqrt{x^2+y^2}|$ and use sandwich theorem (inequalites of $\ln$, such as $\ln x\leq x-1<x,~x>0$ didn't help here).

Thanks in advance.

Answer 1

For small $(x,y)$ closed enough to zero, then \begin{align*} |x\ln(x^{2}+y^{2})|&=-|x|\ln(x^{2}+y^{2})\\ &=|x|\ln\left(\dfrac{1}{x^{2}+y^{2}}\right)\\ &\leq C|x|\cdot\dfrac{1}{(x^{2}+y^{2})^{1/4}}\\ &\leq C\cdot\dfrac{(x^{2}+y^{2})^{1/2}}{(x^{2}+y^{2})^{1/4}}\\ &= C(x^{2}+y^{2})^{1/4}\\ &\rightarrow 0. \end{align*}

Answer 2

Since$$ \lim_{u \to 0^+} u |\ln u| = 0 $$ and$$ |x \ln\sqrt{\smash[b]{x^2 + y^2}}| \leqslant \sqrt{\smash[b]{x^2 + y^2}} |\ln\sqrt{\smash[b]{x^2 + y^2}}|, \quad \forall (x,y) ≠ (0,0) $$ so$$ \lim_{(x, y)\to (0, 0)} x \ln\sqrt{\smash[b]{x^2 + y^2}} = 0. $$

Answer 3

By polar coordinates $r\to 0$

$$x\ln\sqrt{\smash[b]{x^2+y^2}}=r\cos \theta \ln r \to 0$$

indeed

• $0 \le |r\cos \theta \ln r| \le |r\ln r|$

and

• $r\ln r \to 0$