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First half of the proof from Rudin

Second half of the proof from Rudin

How does (16) follow from (21) and (22)? When you take p->A and q->A, it makes sense to me that the c value required for a < x < c to imply (21) and (22) approaches a, but I can't prove that either. Also after that step, how does (16) follow?

cactus
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1 Answers1

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One thing to understand is that $x\in(a,b)$ so that $\lim_{x\to a}$ is effectively the limit as $x$ approaches $a$ from above. Thus, the condition "if $a<x<c_2$" in the proof actually means for any $x$ sufficiently close to $a$, and the same is true for "if $a<x<c_3$."

Equation (21) thus shows that $f(x)/g(x)<q$ for any $q>A$ and $x$ sufficiently close to $a$. Equation (22) that $f(x)/g(x)>p$ for any $p<A$ and $x$ sufficiently close to $a$. These conditions give us that the limit of $f(x)/g(x)$ exists and equals $A$. The existence of the limit is usually defined with epsilons, i.e., defined to mean, given any $\epsilon>0$, that $|f(x)/g(x)-A|<\epsilon$ for $x$ sufficiently close to $a$, but we can see that this holds by taking $q=A+\epsilon$ and $p=A-\epsilon$ in (21) and (22).

ForgotALot
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