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Why is it wrong to write $$\mathop{\lim}\limits_{x \to \infty}x\left(\frac{1}{x}\sin x-1+\frac{1}{x}\right)=(0k-1+0)\cdot\mathop{\lim}\limits_{x \to \infty}x,$$ where $\lvert k \rvert \le 1$?

And, as an aside, is there an idiom or symbol for compactly representing, in an expression, a number that is always within a range so that "where ..." can be avoided?

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    Your suggestions seem brief enough. I would be more concerned about the reasoning embodied in the sample equation.2011-09-19
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    @André: I'm staring at it but missing where I wen't wrong. It's basically copied from [a comment to an answer to another question I had](http://math.stackexchange.com/questions/64273/using-intermediate-value-arguments-at-limits-rather-than-finding-explicit-bounds/64298#64298). Where did I go wrong?2011-09-19
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    @André: Sorry to be dim: Sample equation here or [there](http://math.stackexchange.com/questions/64273/using-intermediate-value-arguments-at-limits-rather-than-finding-explicit-bounds/64298#64298)? (I want to make sure I've corrected the error here that I made there.)2011-09-19
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    This thread has morphed into a useful (for me anyway) discussion of an error I made in the formulation above. Is there an idiom for rewriting the question and title (as something along the lines of "whats wrong with this formulation") so it better matches the answer I have below? (I think I also have the answer to my original question here in the comments, and I could keep that and treat it parenthetically.)2011-09-19
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    @raxacorico: The comment back then was not considered to be of good form. The changed version is better, but still problematical. The "$k$" for example is misleading, that limit does not exist. Crucially, a limit of a product **cannot in general** be expressed as a product of limits.2011-09-20
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    @André: Yes, that's what I was asking about in my previous comment, above: since my my misformulation of this question is more interesting than the question itself, can I refocus the question on that, to match Brian's answer? (This is a question about proper math.se etiquette)2011-09-20
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    @raxacorico: In general a new question that focuses on the actual problem is best. People are not happy when a carefully written answer is made obsolete by a radical reformulation of the question. In this case, the one posted answer actually addresses a question as it might be revised. But still, the title would need to be completely changed, as would the question (the issue is **not** finding a more compact notation). Sounds like a new question.2011-09-20
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    @André: Yes, certainly in general a new question would be better. Here though I've made it match the answer. (And have preserved the effort put into the comments with an aside.) So I think (hope) I'm not stepping on toes.2011-09-20

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You can’t rewrite that comment this way: $\sin x$ is always between $-1$ and $1$, but it isn’t a constant, which is what you’re implying when you pull it outside the limit.

You could write $$\lim\limits_{x\to\infty}x\left(\frac{1}{x}\sin x - 1 + \frac{1}{x}\right) = (0-1+0)\cdot\lim\limits_{x\to\infty}x,$$

provided that you explained why $\lim\limits_{x\to\infty}\dfrac{\sin x}{x}=0$. For that you really do need to write an explanation, not an equation. In an elementary course you should give more detail rather than less, so it might look something like this:

$\vert \sin x\vert \le 1$ for all real $x$, so $\dfrac{-1}{x} \le \dfrac{\sin x}{x} \le \dfrac{1}{x}$ for all $x>0$, and therefore by the sandwich theorem $$0 = \lim\limits_{x\to\infty}\frac{-1}{x} \le \lim\limits_{x\to\infty}\frac{\sin x}{x} \le \lim\limits_{x\to\infty}\frac{1}{x} = 0$$ and $\lim\limits_{x\to\infty}\dfrac{\sin x}{x}=0$.

In a slightly higher-level course you could simply say that $\lim\limits_{x\to\infty}\dfrac{\sin x}{x}=0$ because the numerator is bounded and the denominator increases without bound.

But it’s just as easy to multiply it out to get $$\lim\limits_{x\to\infty}(\sin x - x + 1)$$ and argue that $0 \le \sin x + 1 \le 2$ for all $x$, so $-x \le \sin x - x + 1 \le 2-x$ for all $x$, and hence (again by the sandwich theorem) the limit is $-\infty$.

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    @Srivatsan: Eye-slip: I was already looking at the $+1$ in the multiplied-out form. Glad you caught that; I’m fixing it now.2011-09-19
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    But *I* still would just do what you prescribed, rather than pulling out an $x$ outside. I guess it just depends on taste. :)2011-09-19
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    @Srivatsan: Yes, so would I. As I used to tell my students, good mathematicians are constructively lazy!2011-09-19
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    I've reformulated the question, so this is now the answer to "the question" and not so much "that comment". Thanks!2011-09-20