Using intermediate value arguments at limits rather than finding explicit bounds Again, I apologize for what looks like a very narrow question. But there's possibly a general principal at work here that I'm not grasping.
I understand the answer provided for exercise 3 in chapter 7 of Spivak's Calculus (4E, p.130), but wonder if another approach might work (and be closer to the spirit of the chapter). For example for (ii), to show that $$\sin x = x-1$$ has a solution, can't I take $$f(x)=\sin x - x+1$$ and argue that for large $|x|$, $x>0 \Rightarrow f(x)<0$ and $x<0 \Rightarrow f(x)>0$, so that I can use Theorem 1 (p. 122: $f(x)$ continuous on $[a,b]$ and $f(a)<0<f(b) \Rightarrow \exists x \in [a,b] : f(x)=0$ )?
 A: I am currently teaching a course from Spivak's Calculus, and I think your solution to this problem is entirely correct.
You considered the auxiliary function $f(x) = \sin x - x + 1$ and showed that it is continuous and takes negative values for sufficiently large, positive $x$ and also positive values for sufficiently large, negative $x$.  So by IVT it must take on the value $0$.
You can be a little more explicit about the sufficiently large business though.  For instance, you know that $-1 \leq \sin x \leq 1$ for all $x$, so 
$-x \leq f(x) \leq -x + 2$.
From this one sees that $f(x)$ is non-negative for all $x \leq 0$ and non-positive for all $x \geq 2$.  In this sort of problem, the more complicated the function gets, the more you want to call on "general principles" in order to give you the estimates you need.  For instance, if you had a very complicated polynomial $p(x)$ of degree $19$ in place of $x$, you probably don't want to give explicit values but just use the fact that as $x$ approaches $\pm \infty$, so does $p(x)$, while $\sin x$ stays bounded.
Final comment: to be sure, you don't have to find explicit values for this problem.  But in order to be best understood you should probably say something in the way of justification of what happens for sufficiently large $x$.  Note that in the above paragraph I gave a less explicit answer for a more general class of problems, and as you know there are other problems in the text which are like the one I made up above.  But -- and this is an issue of effective mathematical writing and communication rather than mathematical correctness -- there is a sort of principle of economy at work here.  In order to be best understood, it's generally a good idea to use the simplest arguments you can think of that justify a given claim, and a lot of people find more explicit arguments to be simpler than less explicit ones.  Anyway, not here but sometimes you do have to be explicit and concrete, so it's a good idea to cultivate the ability to do so...
A: Your approach is correct, but the proof is not water-tight. By which I mean that to complete the proof, one needs the following additional proposition: 

Suppose a continuous function $f : \mathbb R \to \mathbb R$ satisfies $f(x) \to \infty$ as $x \to \infty$, and $f(x) \to -\infty$ as $x \to -\infty$. Then there exists some $x_0 \in \mathbb R$ such that $f(x_0) = 0$. 

From your question and comments, I feel that you intuitively understand the above statement, and in fact, you are also implicitly using it. However, I am not sure that you know how to prove it. The proof is just a single line, by the way. Nevertheless, I encourage you to do it. 
I am emphasizing this point because when one is learning a subject, it is of great pedagogical value to sit down and prove even seemingly obvious statements. Moreover, while one certainly cannot write completely rigorous proofs all the time, one should at least be aware of the little missing details, and also how those details can be taken care of. 
A: Perhaps I'm missing something (given the comments posted thus far), but I don't see what is wrong with your argument, other than being a little more precise in applying Theorem 1, such as:
$$f(3) = \sin(3) - 3 + 1 \leq 1 - 3 + 1 = -1 < 0$$
and
$$f(-1) = -\sin(1) - (-1) + 1 \geq -1 + 1 + 1 = 1 > 0,$$
so there exists $x \in [-1,3]$ such that $f(x) = 0$.
