Tricky limit on the open first quadrant: $\lim_{(x,y) \rightarrow (0,0)} -x\log{y} $.

Question

Consider $ f:(0,1) \times (0,1) \rightarrow \mathbb{R} $ where$$ f(x,y)=-x\log{y} $$ I am trying to prove whether $$ \lim_{(x,y) \rightarrow (0,0)}{f(x,y)}=0$$ My current idea is as follows:

Let $ p:(0,1) \rightarrow (0,1) $ be an arbitrary path such that $$ \lim_{x \rightarrow 0}{p(x)}=0 $$ By Taylor's Theorem $$ p(x) = \sum_{i=0}^{\infty}{a_{i}x^{i}} $$ By the criteria for the path, we know that $ a_{0} = 0 $. Also, let $ n \in \mathbb{Z}_{+} $ be the smallest positive integer such that $a_{n} \neq 0$. Furthermore, we know $a_{n} > 0$, again by the definition of $p$.

As $x \rightarrow 0$, $p(x) \approx a_{n}x^{n}$. We can now evaluate $$ \lim_{x \rightarrow 0}{-x\log{p(x)}}$$ $$ = \lim_{x \rightarrow 0}{-x\log{a_{n}x^{n}}}$$ $$ = \lim_{x \rightarrow 0}{[-x\log{a_{n}}-x\log{x^{n}}]}$$ $$ = \lim_{x \rightarrow 0}{-nx\log{x}}= 0$$ I am uncertain whether this approach is rigorous and whether I am allowed to use Taylor's Theorem for the path. Any help is greatly appreciated.

Answer 1

Actually the OP limit does not exist.

Indeed, if there existed $\lim\limits_{(x,y)\to(0,0)}-x\log y=l\;,\;$ then, for any $\,k\in\Bbb R^+\,,\,$ we would consider the path $\;p_k:(0,1)\to(0,1)\,$ defined as $\;p_k(x)=e^{-k/x}\;$ and , $\;$since $\,\lim\limits_{x\to0^+}p_k(x)=0^+\,,\,$ it would also exist the limit

$\lim\limits_{x\to0^+}-x\log p_k(x)=l\;,\;$

but , $\;-x\log p_k(x)=k\,,\;$ for any $\;x\in(0,1)\;,$

consequently,

$k=l\,,\;$ for any $\;k\in\Bbb R^+\;,$

which is a contradiction,

hence the OP does not exist.

Answer 2

Consider changing variables to $z=-\log y$. Note $y \to 0^+$ if and only if $z \to +\infty$. Then $$ \lim\limits_{(x,y)\to(0,0)}-x\log y= \lim\limits_{(x,z)\to(0,+\infty)} xz $$ This has indeterminate form $0\cdot\infty$; the limit does not exist.

Answer 3

Another way

• for $x=y=t\to 0^+$

$$-x\log{y}=-t\log (t) \to 0$$

• for $x=-\frac1{\log(t)}$ and $y=t\to 0^+$

$$-x\log{y}=\frac1{\log(t)}\log (t)=1$$