Tutorial02-solutions(1)

# Note that we have x l 0 so that x l 0 x l therefore

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Unformatted text preview: it follows from the deﬁnition that lim x→a x→a (x) − l = 0 means that for any given &gt; 0, there exists a δ &gt; 0 such that for any 0 &lt; x − a &lt; δ , we have | (x) − l − 0| &lt; . Note that we have (x) − l ≥ 0 so that | (x) − l − 0| = (x) − l . Therefore, we see that for any &gt; 0, there exists a δ &gt; 0 such that for any 0 &lt; x − a &lt; δ , we have (x) − l &lt; , which implies that lim (x) = l by deﬁnition. :) x→a 2 Questions on calculation 22 E Prove that lim ( )→(0 0) 22 + ( − )2 is not deﬁned (by ﬁnding limits along dif- ferent straight lines). S Let’s ﬁnd the limits along the lines = 0 and = . For = 0, we have 22 lim ( →0 lim ( For 0) = lim ) = lim →0 2 02 0 =0 + ( − 0)2 = , we have 2 →0 →0 2 2 2 + ( − )2 =1 Since limits along diﬀerent lines are diﬀerent, the limit of the whole function is not deﬁned. :) E Find the limit if it exists, or show that the limit does not exist. 2 (a) lim )→(2 1) ( 4− 2+3 2 (b) 4 lim )→(0 0) ( 4 +3 2 4 + (c) lim ( )→(0 0) +2 (d) 2 + 2+ )→(0 0 0) ( + 2+ 3 lim 2 3 4 (e) lim )→(0 0) ( S + 2+ 2 (a) ( ) = 4− 2 is a rational function and as such it is continuous every2 +3 where where the denominator is nonzero. In particular, this function is continuous at (2 1), so by continuity lim ( ( )→(2 1) ) = (2 1) = 2 4−2·1...
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