# Finding $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$

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 $$x,y = {2\over n},{1 \over n}$$

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}

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 $$y=\sqrt{2}x$$. 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 $$r=0$$.

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

Substitute $$x= r \cos \theta$$ and $$y= r \sin \theta$$, then you have

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

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

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

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

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 $$(0,0)$$