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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?

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    $\begingroup$ Sounds like its just wrong, clearly you have a sequence on which the limit does not exist, so the whole limit cannot. No idea how it does limits, possibly a truncated taylor series which it doesn't pay enough attention to radii of convergence? $\endgroup$ – Adam Hughes Apr 29 '16 at 22:29
  • $\begingroup$ Just to point it out, the question isn't whether the limit exists or not, but rather why does WA think it exists. Edit: Not only does the limit not exist, but also it doesn't even make sense to question it because limits at infinity are only defined when the domain of the function contains $[a,+\infty[$, for some real $a$ and, as the OP pointed out, this isn't the case. $\endgroup$ – Git Gud Apr 29 '16 at 22:32
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    $\begingroup$ If you try to do it numerically then most likely you won't actually hit $n\pi$ so it will, numerically, appear to converge to $1$. $\endgroup$ – Jared Apr 29 '16 at 22:37
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    $\begingroup$ Maple 17 returns "undefined". Best possible answer. (Mathematica returning the original limit might mean the limit doesn't exist, or that it doesn't know how to do the limit.) $\endgroup$ – Christopher Carl Heckman Apr 29 '16 at 23:08
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    $\begingroup$ Maybe Wolfram has a sense of humor: If you can't tell when it's bleepin' obvious that an expression is unbounded in every neighborhood of $a,$ it returns $1$ as the limit of the expression as $x\to a.$ $\endgroup$ – zhw. Apr 29 '16 at 23:29
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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.

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