# Why is it wrong to express $\mathop{\lim}\limits_{x \to \infty}x\sin x$ as $k\mathop{\lim}\limits_{x \to \infty}x$; $\lvert k \rvert \le 1$?

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. – André Nicolas Sep 19 '11 at 22:27
@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. Where did I go wrong? – raxacoricofallapatorius Sep 19 '11 at 22:34
@André: Sorry to be dim: Sample equation here or there? (I want to make sure I've corrected the error here that I made there.) – raxacoricofallapatorius Sep 19 '11 at 22:59
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.) – raxacoricofallapatorius Sep 19 '11 at 23:45
@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. – André Nicolas Sep 20 '11 at 0:18

## 1 Answer

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. – Brian M. Scott Sep 19 '11 at 23:30
But I still would just do what you prescribed, rather than pulling out an $x$ outside. I guess it just depends on taste. :) – Srivatsan Sep 19 '11 at 23:34
@Srivatsan: Yes, so would I. As I used to tell my students, good mathematicians are constructively lazy! – Brian M. Scott Sep 19 '11 at 23:37
I've reformulated the question, so this is now the answer to "the question" and not so much "that comment". Thanks! – raxacoricofallapatorius Sep 20 '11 at 0:55