A limit involving $\cot$ that seemingly shouldn't exist According to Wolfram Alpha, $$\lim_{x \to \infty} \frac{x - \cot x}{x} =1.$$
But does the limit even exist? Isn't $\frac{x - \cot x}{x}$ unbounded near $x= n \pi$ for all $n \in \mathbb{N}$?
Assuming that the limit doesn't actually exist, what might explain why Wolfram Alpha thinks that it does exist?
 A: When we ask Mathematica (10.4) to compute a power series for $f(x) = \frac{x - \cot(x)}{x}$ around $x = \infty$ (of order 10), we get $$  1 + \cot(x)\left(\frac{-1}{x} + O\left(x^{-12}\right) \right) + O\left(x^{-11}\right)  \text{.}  $$  If we imagine that cotangent was a very nice function (heh), we'd say it is getting crushed by $-1/x$ as $x \rightarrow \infty$ and similarly for the big-O residuals.  This just leaves the "$1$" in the limit.  (This property of Mathematica's Series[] function to use some simple transcendental functions rather than expand them into the series is frequently irritating.)
If we ask Mathematica to evalute the limit, it stares at us blankly
In:  Limit[(x - Cot[x])/x, x -> \infty ]

Out:  Limit[(x - Cot[x])/x, x -> \infty ]

If we ask Wolfram Alpha to do expand $f$ in a series around $\infty$, it stares at us blankly
Series[(x - Cot[x])/x,{x,\infty,2}]

(no series expansion available)

So I can't guarantee that Alpha is using the above expansion to arrive at the wrong limit.  But I'd bet a dollar that it does.
