Tutorial02-solutions(1)

# sin and take the limit as s we make the polar

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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...
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## This document was uploaded on 02/10/2014.

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