This preview shows page 1. Sign up to view the full content.
Unformatted text preview: limit is
beginning so that we have for any ﬁxed > 0,
lim →∞ = lim
1+ →∞ 7 1− 1
1+ > 0 at the It’s easy to see that lim = ∞ for any ∞ lim →∞ > 0 so that
1− = lim →∞ 1+ 1
1+ =1 It follows that
lim lim →0+ →∞ E = lim 1 = 1
→0+ 1+ Find +
lim )→(∞ ∞) ( − 2 2 + = cos θ , Here you can do the polar substitution
by applying the Squeeze Theorem. = sin θ and take the limit as S
We make the polar substitution = cos θ ,
(∞ ∞) means that (
)
+∞ which is precisely
−
+2
cos θ + sin θ
=0
= lim
→∞ (1 − sin(2θ )/ 2)
( )→(∞ ∞) = lim 2 = sin θ and note that (
+∞. It follows that →∞
)→ (cos θ + sin θ )
2 (1 − sin θ cos θ ) +
lim →∞ as 1 − sin(2θ )/2 > 0 for any θ .
E :) :) Find
2 lim )→(∞ ∞) ( + 2 −− You can also do it by polar substitution.
S
We make...
View
Full
Document
This document was uploaded on 02/10/2014.
 Spring '13

Click to edit the document details