A sequence defined through an arc tangent and arc cotangent Let $$f:\mathbb{R^*}\to \mathbb{R}, f(x) = \arctan{\dfrac{1}{x}}-\operatorname{arccot}{\dfrac{1}{x}}$$
I found that this function is decreasing on $\mathbb{R^*}$ and a bijection for the same domain and for the range $(-\dfrac{3
\pi}{2};\dfrac{\pi}{2})$.
If we define $x_n$ the sequence of solutions to the equation $$f(x)=\dfrac{1}{n}$$ than prove that $x_n$ is convergent and find its limit.
Here is what I did:
$$\arctan\dfrac{1}{x_n}=\dfrac{1}{n}+\operatorname{arccot}\dfrac{1}{x_n}$$
$$\dfrac{1}{x_n}=\tan \left( \dfrac{1}{n}+\operatorname{arccot}\dfrac{1}{x_n} \right)$$
$$\dfrac{1}{x_n}=\dfrac{\tan\dfrac{1}{n}+x_n}{1-x_n\tan\dfrac{1}{n}}$$
After recombination, we have:
$$(x_n)_{1,2}=\dfrac{-\sin\dfrac{1}{n}+1}{\cos\dfrac{1}{n}}$$
I have already taken the positive solution. For $n$ a natural number both $\sin\dfrac{1}{n}$ and $\cos\dfrac{1}{n}$ are positive. 
Obviously the limit of the sequence is $1$, but how do I proove the convergence? Thank you!
 A: $f(x) = \arctan{\frac{1}{x}}-arccot{\frac{1}{x}}$
Since
$\arctan(x)+arccot(x)
=\frac{\pi}{2}
$,
this becomes
$f(x) 
= 2\arctan{\frac{1}{x}}-\frac{\pi}{2}
$.
If
$f(x_n) = \frac1{n}$.
then
$2\arctan{\frac{1}{x_n}}
=\frac{\pi}{2}+\frac1{n}
$
or
$\begin{array}\\
\frac{1}{x_n}
&=\tan(\frac{\pi}{4}+\frac1{2n})\\
&=\frac{\tan(\frac{\pi}{4})+\tan(\frac1{2n})}
{1-\tan(\frac{\pi}{4})\tan(\frac1{2n})}\\
&=\frac{1+\tan(\frac1{2n})}
{1-\tan(\frac1{2n})}\\
\text{so}\\
x_n
&=\frac{1-\tan(\frac1{2n})}{1+\tan(\frac1{2n})}\\
&=1-\frac{2\tan(\frac1{2n})}{1+\tan(\frac1{2n})}\\
&=1-\frac{2(\frac1{2n}+(\frac1{n^2}))}{1+\frac1{2n}+(\frac1{n^2})}\\
&=1-\frac1{n}(1+\frac1{n})(1-\frac1{2n}+(\frac1{n^2}))\\
&=1-\frac1{n}(1+\frac1{2n}+\frac1{n^2})\\
&\to 1\\
\end{array}
$
A: HINT: Prove that, as $n \to \infty$,
$$\frac{1}{n}\to 0$$
which means that $x_n$ approaches the solution to the equation
$$\arctan \frac{1}{x_n}-\operatorname{arccot} \frac{1}{x_n}=0$$
And show that the unique solution to this equation is $x_n=1$.
