Spivak's Calculus 19-25 This question is from Spivak's Calculus (3rd ed) 19-25:

I'm a bit puzzled by the (partial) solution:

I'm not sure how to interpret the integral with the limit in front of it.  Nor do I understand how we get equality from $\phi_h(x)=0$.
According to the problem we can't use part (b) to answer part (d).  Does anyone have a simple approach to part (d)?
 A: Hint. Let $h>0$, then  $\phi_h(x)=\frac{\phi(x/h)}{h}=0$ if $|x/h|\geq 1$, that is for $|x|\geq h$. Therefore, if $0<|h|\leq 1$ then 
$$\int_{-1}^1 \phi_h(x)f(x)\,dx=\int_{-h}^h \phi_h(x)f(x)\,dx.$$
Moreover, let $\epsilon>0$, then, by the continuity of $f$ at $0$, there is $\delta>0$ such that for any $x$ s.t. $|x|<\delta$,  $-\epsilon<f(x)-f(0)<\epsilon$. Hence for $0<h<\delta$, by using (a),
$$\int_{-h}^h \phi_h(x)f(x)\,dx-f(0)=\int_{-h}^h \phi_h(x)(f(x)-f(0))\,dx<\epsilon\int_{-h}^h \phi_h(x)=\epsilon$$
Can you take it from here?
A: Specifically addressing your concern, recall that by assumption $\phi(y)=0$ for any $|y|\geq 1$. Then setting $y=\frac{x}{h}$, we have 
$$
\phi(y)=\phi(x/h)=0
$$
whenever $|y|=\left|\frac{x}{h} \right|\geq 1\implies |x|\geq h$
Then for $h<\delta<1$, we have 
$$
\int_{-1}^1\phi_h(x)\mathrm dx=\int_{h}^1\phi_h(x)\mathrm dx+\int_{-h}^h\phi_h(x)\mathrm dx+\int_{-1}^{-h}\phi_h(x)\mathrm dx\\
=\int_{-h}^h\phi_h(x)\mathrm dx
$$
since the two outer integrals are zero.
