Tutorial02-solutions(1)

# In other words we either must have 0 or 0 thus the

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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.

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