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I'm wondering what the most graceful/accurate way to state the limit of $\cos(x) - 1$ as x approaches $-\infty$ doesn't exist. Intuitively, this limit can't exist; but I'm unsure if using the fact that all subsequences of a sequence must converge for the limit to exist would work here (My original approach) as this seems more like the limit of a function rather than a sequence (Even when you define a sequence as having elements equivalent to this).

Could I get some opinions/headings?

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  • $\begingroup$ Using the sequential definition of the limit, if your function has a limit in $-\infty$ then for every sequence having a limit in $-\infty$ then, the limit should be the same. So you can still use a subsequence to prove your argument. $\endgroup$ – Oussama Boussif May 17 '17 at 16:33
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Hint:

Check the sequences $${x_n}=-\frac{2n+1}{2}\pi$$ and $${y_n}=-4n\pi$$, converging to negative infinity. Then put them in the function and find it violates seqential definition of limit of function.

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If you wish to use an $\epsilon$ argument consider the following

  1. $-2\le \cos(x) -1\le0$ for all $x\in\mathbb{R}$

  2. Assume $\lim_{x\to-\infty}(\cos(x) -1)=L$

  3. Then there is an $x_0\in(-\infty,0)$ such that if $x<x_0$ then

$$ -\frac{1}{2}<\cos(x)-1-L <\frac{1}{2}$$

so for all $x<x_0$ it is the case that

$$L-\frac{1}{2}<\cos(x)-1<L+\frac{1}{2}$$

  1. But since the interval $\left(L-\frac{1}{2},L+\frac{1}{2}\right)$ has length $1$ and since $\cos(x)-1$ is a periodic function with codomain $[-2,0]$ of length $2$ there must exist an $x<x_0$ lying outside the interval $\left(L-\frac{1}{2},L+\frac{1}{2}\right)$. For such an $x$, the inequality in $(3)$ fails.

  2. This contradicts the assumption in step $(2)$.

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