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?
 A: 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 $-\infty$.
