$$\lim_{(x,y)\to (0,0)} \frac{\log \left(|x| + e^{|y|}\right)}{\sqrt{x^2 + y^2}} = ?$$

Assuming that $\log \triangleq \ln$, then I tried the following:

1. Sandwich rule

Saying that $\log \left(|x| + e^{|y|}\right) < |x| + e^{|y|}$: $$\begin{align} \lim_{(x,y)\to (0,0)} \frac{\log \left(|x| + e^{|y|}\right)}{\sqrt{x^2 + y^2}} &< \\ & \lim_{(x,y)\to (0,0)} \frac{|x| + e^{|y|}}{\sqrt{x^2 + y^2}} = \\ &= \lim_{r \to 0} \frac{|r\cos \theta| + e^{|r\sin\theta|}}{r} \\ &= \lim_{r \to 0} |\cos\theta| + \frac{e^{|r\sin\theta|}}{r} \end{align}$$ From here it seems that the limit doesn't exist, so it doesn't indicate anything on the given function.

2. Polar coordinates

Tried expressing $x=r\cos\theta, y=r\sin\theta$, though got stuck right at the $\log$ function.
Also tried using it in the Sandwich rule above, to no avail.

3. Single variable assignment

Another technique is to replace an expression of $x$ and $y$ with a single variable $t$, but for this case it is not helpful.

The $\sqrt{x^2 + y^2}$ strongly indicates on Polar, though I can't work through that $\log$ and $e$.
It seems that I'm missing an important logarithmic identity, though I've seen many identities at Wiki and none is useful.

  • 2
    $\begingroup$ Hint: The limit of the function at $(x,0)$ when $x\to0$ is $1$. The limit of the function at $(x,x)$ when $x\to0$ is $\sqrt2$. QED. $\endgroup$
    – Did
    Mar 19, 2015 at 20:56
  • $\begingroup$ @Did: Why are we allowed to set $y$ to 0? The parameter $y$ should approach 0... not to be equal to it. $\endgroup$
    – Dor
    Mar 19, 2015 at 22:12
  • 1
    $\begingroup$ No, it is the parameter (x,y) that should approach (0,0) without being equal to (0,0). The parameter y can very well be 0. // Anyway, the limit along (x,2x) is 3/sqrt(5), again different from the limit sqrt(2) along (x,x), hence the choice of a contradiction should not be a problem. $\endgroup$
    – Did
    Mar 20, 2015 at 9:51

1 Answer 1


This is a very non-rigorous approach, but intuitive.

The Taylor expansion (about $v=1$) of $\log(v)$ is $(v-1)+O(v^2)$, and the Taylor expansion (about $u=0$) of $e^u$ is $1+u+O(u^2)$.

Thus $\log(|x|+e^{|y|})$ is approximately $|x|+e^{|y|}-1$ which is approximately $|x|+|y|+O(x^2)+O(y^2)$. In polar coordinates this is $|r|(|\cos \theta|+| \sin \theta|) + O(r^2)$.

Now as $r \to 0$, clearly the value of $\frac{|r|(|\cos \theta| +|\sin \theta|)}{|r|}=|\cos \theta| + |\sin \theta|$ will vary with $\theta$, so the limit cannot be well-defined. Someone ought to check this though, I don't know yet if it's correct.


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