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

Answer

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}\end{eqnarray*}