TITLE: Using intermediate value arguments at limits rather than finding explicit bounds
QUESTION [3 upvotes]: 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 $f(a)<0<f(b) \Rightarrow \exists x \in [a,b] : f(x)=0$ )?

REPLY [2 votes]: 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.
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