
#1
May211, 08:16 PM

P: 6

When I try to demonstrate that lim (x^2)=4
x>2 I found a different delta (delta=min{2sqrt(epsilon4),sqrt(epsilon+4)2}, towards the one that is written in Demidovich´s book (delta=epsilon/5). Could someone help me? Could someone tell me, too, a good algebra, calculus and real analysis book? Is it ok if i try analysis next semester absent a more advanced calculus (without calculus 2)? I need some books that explain polar coordinates too (good ones for beginner). Thanks and sorry for my mathematical notation. 



#2
May311, 08:12 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

You want [itex]x^2 4= (x 2)(x+ 2)= x+ 2x 2< \epsilon[/itex].
So we can write [itex]x 2< \epsilon/x+ 2[/itex] But we need a value, [itex]\delta[/itex] that does NOT depend on x so that is [itex]x 2< \delta[/itex] then [itex]x 2< \epsilon/x+ 2[/itex]. That means we want [itex]\delta< \epsilon/x+2[/itex]. If x 2< 1 (The "1" is chosen just because it is simple. Any positive number would do.) then 1< x 2< 1 so 1< x< 3 and then 3< x+2< 5 so that 3< x+ 2< 5. Then [tex]\frac{1}{5}< \frac{1}{x+2}< 1/3[/tex] and so [tex]\frac{\epsilon}{5}< \frac{\epsilon}{x+2}[/tex] That is, if [tex]x 2< \frac{\epsilon}{5}[/tex] and x 2< 1 as assumed above to get this, we will have [tex]x2< \frac{\epsilon}{x+2}[/tex] so [tex]x2x+2= x^2 4< \epsilon[/tex] as need. In other words, take [itex]\delta[/itex] to be the smaller of [itex]\epsilon/5[/itex] or 1. Analysis is basically the theory behind the Calculus. I would NOT take an analysis course without having completed the Calculus sequence. 


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