Question about existence of $\lim\limits_{x\to +\infty} e^{-x}\left(1+\frac{1}{1+\cot x}\right)$ I read on a book that $\displaystyle\lim_{x\to +\infty} e^{-x} \left(1+\frac{1}{1+\cot x}\right)$ does not exist. I tried to find out why, by choosing sequences $x_n$ and $y_n$ such that $\cot x_n=0$ and $\cot y_n=1$, $\forall~ n\in\mathbb{N}.$ However, we have the factor $e^{-x}$ as a result the limit goes to $0$ over those two sequences. Any Ideas would be helpful. Thanks a lot.
 A: For any $\;n\in\Bbb N\;,\;$ we consider the following function :
$f(x):\left]n\pi-\dfrac{\pi}4,\,n\pi\right[\to\Bbb R\;$ defined as
$f(x)=e^{-x}\left(1+\dfrac1{1+\cot x}\right)\quad$ for all $\;x\in\left]n\pi-\dfrac{\pi}4,\,n\pi\right[.$
The function $\,f(x)\,$ is continuous on $\left]n\pi-\dfrac{\pi}4,\,n\pi\right[,$ moreover
$\lim\limits_{x\to\left(n\pi-\frac{\pi}4\right)^+} f(x)=-\infty\;\;$ and $\;\lim\limits_{x\to n\pi^-}f(x)=e^{-n\pi}>0\;.$
By applying the intermediate value theorem, it follows that there exist $\;a_n,\;b_n\in\left]n\pi-\dfrac{\pi}4,\,n\pi\right[\;$ such that $\;f(a_n)=-1\;$ and $\;f(b_n)=-2\;.$
Consequently, there exist two sequence $\,\big\{a_n\big\}_{n\in\Bbb N}\;$,$\,\big\{b_n\big\}_{n\in\Bbb N}$ such that $\;\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}b_n=+\infty\;\;$ and $\;f(a_n)=-1\;,\;f(b_n)=-2\;,\quad\forall\,n\in\Bbb N\,.$
If there existed the limit $\;\lim\limits_{x\to+\infty}e^{-x}\left(1+\dfrac1{1+\cot x}\right)=l\;\;,$
it would result that
$\lim\limits_{n\to\infty}e^{-a_n}\left(1+\dfrac1{1+\cot a_n}\right)=\lim\limits_{n\to\infty}\big(\!-1\big)=l\quad$ and
$\lim\limits_{n\to\infty}e^{-b_n}\left(1+\dfrac1{1+\cot b_n}\right)=\lim\limits_{n\to\infty}\big(\!-2\big)=l\;\;,$ consequently,
$l=-1=-2\;$ which would be a contradiction.
Hence, it does not exist the limit $\;\lim\limits_{x\to+\infty}e^{-x}\left(1+\dfrac1{1+\cot x}\right).$
