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Unformatted text preview: ays deﬁned, but for ln( ) we need
> 0. In other words, we either must have
> 0 or
< 0. Thus the domain
of this map is the union of the open I and III quadrants.
:) 1 E :R Given a map lim
x→a → R , prove that lim (x) = l holds if and only if
x→a (x) − l = 0 does.
We ﬁrst prove the “only if" part. For this, we assume lim (x) = l holds and S x→a we want to show this implies that lim
x→a (x) − l = 0. Now it follows from the deﬁnition
> 0, there exists a δ > 0 such that for that lim (x) = l means that for any given
x→a any 0 <
(x) − l
such that
that lim
x→a x − a < δ , we have (x) − l < . Note that we have (x) − l ≥ 0 so that
=  (x) − l − 0. Therefore, we see that for any > 0, there exists a δ > 0
for any 0 < x − a < δ , we have  (x) − l − 0 = (x) − l < , which implies
(x) − l = 0 by deﬁnition. We now prove the “if" part. For this, we assume lim
x→a (x) − l = 0 holds and we want to show this implies that lim (x) = l. Now...
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This document was uploaded on 02/10/2014.
 Spring '13

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