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Find bounds for $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$

I am finding maximum and minimum for function and one of critical case is to find possible minimal and maximal value of given function in $0,0$. But how can I do this due to this limit doesn't exists (for example we can take $x,y = {1\over n},{1\over n}$ and $x,y = {2\over n},{1 \over n}$

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You may use AM-GM (inequality between arithmetic and geometric mean) and the mean value theorem to get reasonable bounds:

  • AM-GM: $a+b \geq 2\sqrt{ab}$ with $a =2x^2, b= y^2$ and equality iff $a=b \Leftrightarrow 2x^2 = y^2$
  • MVT: For $y \neq 0, y>-1$ you have $\frac{\ln (1+y)}{y} = \frac{1}{1+\eta}$ with $\eta$ between $0$ and $y$

Consider $(x,y)$ with $xy\neq 0$ (otherwise the expression is equal to $0$ anyways). For convenience assume further $|x|,|y| < 1$, since we want to study the behaviour of the expression around $(0,0)$:

\begin{eqnarray*}\left| \frac{x \ln(1+y)}{2x^2+y^2}\right| & \stackrel{AM-GM}{\leq} & \frac{|x||\ln(1+y)|}{2\sqrt{2x^2y^2}} \\ & = & \frac{1}{2\sqrt{2}}\cdot \left| \frac{\ln(1+y)}{y}\right| \\ & \stackrel{MVT}{=} & \frac{1}{2\sqrt{2}}\cdot\frac{1}{1+\eta} \\ & \leq & \begin{cases}\frac{1}{2\sqrt{2}}\cdot\frac{1}{1+0} & y>\eta> 0 \\ \frac{1}{2\sqrt{2}}\cdot\frac{1}{1-|y|} & -1 < y < 0 \end{cases}\\ & \stackrel{e.g. \color{blue}{|y|<\frac{1}{2}}}{\leq} & \begin{cases}\frac{1}{2\sqrt{2}} & y> 0 \\ \frac{1}{\sqrt{2}} & \color{blue}{-\frac{1}{2}<y<0} \end{cases}\\ \end{eqnarray*}

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Every level curve

$$ f(x,y)= \frac{x \ln(1+y)}{2x^2+y^2}=c $$

for $c\le0.353$ passes through $(0,0)$ but not so for $c>0.354$, so the maximal point $(0,0,z_{\text{MAX}})$ should have $z_{\text{MAX}}$ lie somewhere between those two quantities.

For each level curve, the point nearest $(0,0)$ appears to lie along the line $y=\sqrt{2}x$. Replacing $y$ with $\sqrt{2}x$ gives

$$ y=\frac{\ln(1+\sqrt{2}x)}{4x} $$

and

$$ \lim_{x\to0}\frac{\ln(1+\sqrt{2}x)}{4x}=\frac{\sqrt{2}}{4}\approx0.3535533906 $$

This would be the correct maximum if indeed the point on each level curve closest to $(0,0)$ actually lies along the line $y=\sqrt{2}x$.

level curve

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You're right that $\lim_{(x,y)\rightarrow(0,0)}$ doesn't exist. What you can do is make a substitution $x=r\cos\phi$, $y=r\sin\phi$, $r\in(0,\infty)$, $\phi \in [0,2\pi)$ and look for the minimum of function $$ \frac{r\cos\phi \ln(1+r\sin\phi)}{2r^2\cos^2\phi+r^2\sin^2\phi}$$ Note that (using the de l'Hospital rule, or squeezing $\frac{x}{1+x} \le \ln x \le x$) you can find that $$\lim_{r\rightarrow 0} \frac{r\cos\phi \ln(1+r\sin\phi)}{2r^2\cos^2\phi+r^2\sin^2\phi} = \frac{\cos\phi \sin\phi}{2\cos^2\phi+\sin^2\phi}$$ which is a well defined function for all $\phi\in[0,2\pi)$, which allows you to analyze the function close to $r=0$.

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  • $\begingroup$ How did that very last equality happen? $\endgroup$ – DonAntonio Jun 9 '19 at 18:35
  • $\begingroup$ @DonAntonio It can be calculated from the de l'Hospital rule like Tesla attempted, of from the squeezing $$ \frac{x}{1+x} \le -\ln(1-\frac{x}{1+x}) = \ln(1+x) \le x$$ $\endgroup$ – Adam Latosiński Jun 9 '19 at 18:39
  • $\begingroup$ Oh, I agree...yet it could probably be a good idea you explicitly write that there to make your intention clear. $\endgroup$ – DonAntonio Jun 9 '19 at 18:40
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    $\begingroup$ Perhaps it'd be a good idea to observe that $$\left|\frac{\cos\theta\sin\theta}{2\cos^2\theta+\sin^2\theta}\right|=\left|\frac12\frac{\sin2\theta}{\cos^2\theta+2}\right|\le\frac{|\sin2\theta|}4$$ $\endgroup$ – DonAntonio Jun 9 '19 at 18:44
  • $\begingroup$ @DonAntonio Sure, but I wanted to leave something to do for the OP, only solving the problem fully if they need further help. $\endgroup$ – Adam Latosiński Jun 9 '19 at 18:48
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Substitute $x= r \cos \theta$ and $y= r \sin \theta$, then you have

$$\lim_{r \to 0}\frac{r \cos \theta \ln(1+r \sin \theta)}{2 r^2 \cos ^2 \theta + r^2 \sin^2 \theta}=\lim_{r \to 0}\frac{ \cos \theta \ln(1+r \sin \theta)}{2 r \cos ^2 \theta + r \sin^2 \theta}\stackrel{l'Hospital}{=}\lim_{r \to 0}\frac{\frac{\cos \theta \sin\theta}{1+r \sin \theta}}{2 \cos^2\theta + \sin^2 \theta}=\lim_{r \to 0}\frac{\cos \theta\sin \theta}{(2 \cos^2 \theta + \sin^2\theta)(1+r \sin \theta)}=\frac{\cos \theta\sin \theta}{2\cos^2 \theta+\sin^2 \theta}.$$

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    $\begingroup$ You didn't correctly calculate the derivative of the denominator; you should have $2\cos^2\theta+\sin^2\theta$ instead of just $2\cos^2\theta$. $\endgroup$ – Adam Latosiński Jun 9 '19 at 18:37
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    $\begingroup$ Even if you used L'Hospital in that second equality sign, the denominator's derivative is wrong. $\endgroup$ – DonAntonio Jun 9 '19 at 18:37
  • $\begingroup$ Yea I didn't see taht I still had an $r$ there, edited it. So am I allowed to use L'Hopital? $\endgroup$ – Tesla Jun 9 '19 at 18:42
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    $\begingroup$ Yeah, you can use L'Hopital. You still have an error in the denominator - the cosinus should be squared. $\endgroup$ – Adam Latosiński Jun 9 '19 at 18:43
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    $\begingroup$ These bounds are wrong. For example for $\cos \theta = \sqrt{\frac13}$, $\sin\theta = \sqrt{\frac23}$ you have $$ \frac{\cos\theta\sin\theta}{2\cos^2\theta+\sin^2\theta} = \frac{\sqrt{2}/3}{4/3} = \frac{1}{2\sqrt{2}} > \frac13$$ $\endgroup$ – Adam Latosiński Jun 9 '19 at 21:46
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Attempt:

$|\log (1+y)| \le 2|y|$, for $|y|\le 1/2$.

Then

$ \dfrac{|x\log (1+y)|}{2x^2+y^2} \le \dfrac{2|xy|}{2x^2+y^2} \le $

$\dfrac{x^2+y^2}{(2x^2+y^2)}< 1$.

Hence for $|y| \le 1/2$:

$|\dfrac{x\log (1+y)}{2x^2+y^2}| <1$.

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Let's have $f(x,y)=\dfrac{x\ln(1+y)}{2x^2+y^2}$

For $x\neq 0$ we have $f(x,0)=0\to 0$

On the other hand $f(x,x)=\dfrac{x\ln(1+x)}{3x^2}=\dfrac{\ln(1+x)}{3x}\sim\dfrac x{3x}\to \dfrac 13$

Since $f$ has at least two different limits along different paths, then $f$ has no limit in $(0,0)$

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