Limit with two-variable function

Question

$f(x,y)=\frac{\ln(x+e^y)}{\sqrt{x^2+y^2}}, \;x\geqslant0,\; x^2+y^2\neq0$. Find the limit when $(x,y) \to (0,0)$. I tried it with Maclaurin and then polar change but can't find it.

Answer 1

Hint: Notice that $$f(0, y) = \frac{\ln(e^y)}{\sqrt{y^2}} = \frac{y}{|y|}.$$

Answer 2

Note that

• $x=t\to 0^+,\,y=0\implies \frac{\ln(x+e^y)}{\sqrt{x^2+y^2}}=\frac{\ln(1+t)}{t}\to 1$

• $x=0,\,y=-t\to 0\implies \frac{\ln(x+e^y)}{\sqrt{x^2+y^2}}=\frac{-t}{t}\to -1$

then the limit doesn't exist.