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I have to check if the function $f(x,y)=|x|\ln(1+y)$ is differentiable at $(0,0)$. I set up the limit: $$\lim_{(x,y)\rightarrow (0,0)}\frac{f(x,y)-f(x_0,y_0)-\frac{\partial f}{\partial x}(x_0,y_0)(x-x_0)-\frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)}{\sqrt{(x-x_0)^{2}+(y-y_0)^{2}}}$$ I calculate from the definition the partial derivatives $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$ and get that both have value $0$ which is also $f(0,0)$. So I have to calculate the limit: $\lim_{(x,y)\rightarrow (0,0)}\frac{|x|\ln(1+y)}{\sqrt{x^{2}+y^{2}}}$. I assumed $\ln(1+y)$~$y$ as $y\rightarrow 0$, so the limit becomes: $\lim_{(x,y)\rightarrow (0,0)}\frac{|x|y}{\sqrt{x^{2}+y^{2}}}$. I now have problems, and tried the inequality: $\frac{|x|y}{\sqrt{x^{2}+y^2}}\le \frac{|xy|}{\sqrt{x^{2}+y^{2}}}\le \frac{|xy|}{|x+y|}$ but it seems not to be the right path. Help is appreciated.

Sine of the Time
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    try with $\frac{|xy|}{\sqrt{x^2+y^2}}\le\frac{|xy|}{\sqrt{x^2}}=|y|$ – Sine of the Time Nov 03 '23 at 20:49
  • @SineoftheTime picky remark, but: that’s not a good way of justifying the inequality, since now you have to make the extra assumption that $x\neq 0$, rather than merely $(x,y)\neq 0$. We can simply jump to $\frac{|xy|}{\sqrt{x^2+y^2}}\leq |y|$, since $|x|=\sqrt{x^2}\leq \sqrt{x^2+y^2}$, so the quotient is bounded by $1$. – peek-a-boo Nov 03 '23 at 20:56
  • @peek-a-boo thank you for your remark. What if first I consider $x\neq 0$ and apply the inequlity and then I assume $y\neq0$ and apply $\frac{|xy|}{\sqrt{x^2+y^2}}\le|x|$ ? Does this work? – Sine of the Time Nov 03 '23 at 21:15
  • Ok, so you have two separate bounds now for different cases, which is just troublesome, especially given that both bounds hold in more generality by a minor rewording of the proof. – peek-a-boo Nov 03 '23 at 21:18
  • @peek-a-boo makes sense. Thank you – Sine of the Time Nov 03 '23 at 21:28

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