Let $g(x,y)$ be a function such that:
I. $-1\lt g(x,y)\lt1.$
II. $$\ln(\frac{1+g(x,y)}{1-g(x,y)})+2y\tan^{-1}(yg(x,y))=2(y^2+1)x,$$ for $x\in\mathbb R, y\gt1.$
Then
i. Show that $g(x,y)$ is increasing in $x.$
ii. Find $$\lim_{y\rightarrow\infty} g(x,y).$$
iii. Show that $g(x,y)$ is differentiable.
iv. Find $\lim_{y\rightarrow\infty}\frac{\partial}{\partial x}g(x,y).$
My question is: how to evaluate the limit maked as ii. ?
I have made an attempt to solve the problem, but I still lack a pice to complete the puzzle.
So it helps a lot if someone can tell me how to evaluate the limit
Thanks in advance.
Best Answer
Since it looks messy to put my efforts in the question body, I decided to write an answer to explain what I have shown and where I got stuck, in an attempt to get an answer. :)
Differentiability of $g$:
Define $f(x,y,z)=\ln(\frac{1+z}{1-z})+2y\tan^{-1}(yz)-2(y^2+1)x.$
Then, for any pair $(x,y)$ with $x\in\mathbb R, y\gt0,$ we find that $\lim_{z\rightarrow1}f(x,y,z)=\infty,$ and $\lim_{z\rightarrow-1}f(x,y,z)=-\infty.$ Hence, by the mean-value theorem, there is $z_0$ such that $-1\lt z_0\lt1$ and $f(x,y,z_0)=0.$ Thus we can apply the implicit-function theorem, and deduce that there is a nbd $U$ of $(x,y),$ and a differentiable function $g_U:U\rightarrow \mathbb R$ such that $g_U$ is the unique solution to the functional equation on $U.$ This shows that $g\mid_U=g_U,$ and hence $g$ is differentiable everywhere, i.e. $g$ is differentiable.
Monotonicity of $g$ in $x$:
By the implicit-function theorem, we find that $$\frac{\partial}{\partial x}g(x,y)=\frac{2(y^2+1)}{\frac{1}{1+g(x,y)}-\frac{1}{1-g(x,y)}+\frac{2y^2}{1+(yg(x,y))^2}}\gt0.$$
This shows that $g(x,y)$ is increasing in $x.$
On the limit $\lim_{y\rightarrow\infty}g(x,y)$ : I rewrite the equation as:
$$2y(\tan^{-1}(yg(x,y))-xy)=2x-\ln(\frac{1+g(x,y)}{1-g(x,y)})$$
$$\tan^{-1}(yg(x,y))=xy+\frac{x}{y}-\frac{\ln(\frac{1+g(x,y)}{1-g(x,y)})}{2y}$$
$$g(x,y)=\frac{tan(xy+\frac{x}{y}-\frac{\ln(\frac{1+g(x,y)}{1-g(x,y)})}{2y})}{y}.$$
But I still cannot infer what the limit should be from this.
The Second Limit
is a direct consequence of the implicit-function theorem and the above limit.
This is all I know about the question, in the dearth of a piece to complete the puzzle.
Hope this helps clarify my question.