Is Spivak Cheating? I'm trying to understand this proof in the Spivak's book Calculus on manifolds page 16:

When I read this I had a feeling that in this part Spivak was informal and intuitive, almost like "cheating" to have a more clean proof. 
Of course, it has to have some formal argument behind, but I couldn't identify it. I think it has to do with limits of composition of functions, where one of the functions is $tx$.
Could anyone clarify this part?
Thanks
 A: You fix $x \neq 0$ and $h=tx$ for $t \to 0$. You have proved that
$$
\lim_{h \to 0} \frac{|\lambda (h)-\mu (h)|}{|h|}=0,
$$
so
$$
\lim_{t \to 0} \frac{|\lambda (tx)-\mu (tx)|}{|tx|}=0
$$
as well. But $\lambda (tx)=t \lambda (x)$ and $\mu (tx)=t \mu (x)$ by linearity, and you conclude.
A: The second chain of inequality shows that $$\lim _{h\to 0} \frac{|\lambda (h) - \mu (h)|}{|h|}=0.$$
A necessary condition for the convergence is that $\phi (t)\to 0$ if $t\to 0$ , where $\phi$ is the quotient calculated in $tx$. It's something like “$a_n \to a$ implies $a_{n_k}\to a$ for each subsequence $a_{n_k}$”.
A: Suppose we know that $\displaystyle \lim_{h \to 0} F(h) = 0$.  This means that for every $\epsilon > 0$, there exists $\delta > 0$ such that
$$|h| < \delta  \implies |F(h)| < \epsilon.$$
Fix $x \in \mathbb{R}^n$.  If $|t| < \delta/|x|$, then $|tx| = |t||x| < \delta$, so $|F(tx)| < \epsilon$.  This says exactly that $\displaystyle \lim_{t \to 0} F(tx) = 0$.
In your situation, take
$$F(h) := \frac{|\lambda(h) - \mu(h)|}{|h|}.$$

Edit: If $\displaystyle \lim_{h \to h_0} F(h) = L$ and $\displaystyle \lim_{t \to t_0} g(t) = h_0$, then $\displaystyle \lim_{t \to t_0} F(g(t)) = L$.
Proof: Fix $\epsilon > 0$.  By the first limit hypothesis, there exists $\delta_1 > 0$ such that
$$|h - h_0| < \delta_1 \implies |F(h) - L| < \epsilon$$
By the second limit hypothesis, there exists $\delta_2 > 0$ such that
$$|t - t_0| < \delta_2 \implies |g(t) - h_0| < \delta_1.$$
Therefore, if $|t - t_0| < \delta_2$, then $|F(g(t)) - L| < \epsilon$.  This says exactly that $\lim_{t \to t_0} F(g(t)) = L$.
Remark: Note that no continuity hypotheses were necessary.  A continuity hypothesis on $F$ would be needed to show instead that $\displaystyle \lim_{t \to t_0}F(g(t)) = F(h_0)$.
A: $\mu$ and $\lambda$ are linear transformations and $t$ is a scalar so
$$
\mu(tx) - \lambda(tx) =t\mu(x) - t\lambda(x) = |t|(\mu(x) - \lambda(x)). 
$$
So 
$$
\left|\mu(tx) - \lambda(tx)\right| = |t|\left|\mu(x) - \lambda(x)\right|.
$$
Also |tx| = |t||x| is a basic property of the euclidean norm, so
$$
\left|\frac{\mu(tx) - \lambda(tx)}{tx}\right|=\frac{|t||\mu(x) - \lambda(x)|}{|t||x|}  = \left|\frac{\mu(x) - \lambda(x)}{x}\right|
$$
for $ t \neq 0$.
Spivak asserts $0 = \lim_{t \to 0} \left|\frac{\mu(tx) - \lambda(tx)}{tx}\right|$. So for all $\epsilon >0$ there exists a $\delta >0$ so that when $0<t<\delta$ 
$$
\epsilon > \left|\left| \frac{\mu(tx) - \lambda(tx)}{tx}\right| - 0\right| = \left| \frac{\mu(tx) - \lambda(tx)}{tx}\right| = \left|\frac{\mu(x) - \lambda(x)}{x}\right|.
$$
Since 
$$ \left|\frac{\mu(x) - \lambda(x)}{x}\right| < \epsilon$$
for all $\epsilon > 0$
$$ \left|\frac{\mu(x) - \lambda(x)}{x}\right|  = 0.$$
Since $|x| > 0$ then $\frac{1}{|x|} > 0$ so we must have
$$ |\mu(x) - \lambda(x)| = 0$$
since there are no nilpotent elements in $\mathbb{R}$
