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2.) compare dominant terms of numerator and denominator. we have in general as x>infinity factorial > exponentials > polynomials/positive powers > linear > logarithmic > constant with polynomials/powers if a>b>0 then x^a > x^b

I did it forwards... you are supposed to find d not e, remember it's epsilon - delta, not delta-epsilon. the way you have it is backwards and if teachers are picky.. they will require you to start with the M and prove |f(x) -L| < e if 0<|x-a|<d

oh.. I inserted the message to underthe sun in the wrong place.. beh.. I need to study :(, mine is coming up 16:00 tomorrow

sun: thought you would post the test questions up :D rage: hmm... let M(e) = 25/e^2 if x > M(e) then: x> 25/e^2 sqrt(x) > 5/e sqrt(x) + 5 > 5/e e > 5/[sqrt(x) + 5] e > | -5 / [ sqrt(x) + 5] | e > | sqrt(x)/[sqrt(x) + 5] -1 | :D ofcourse the hard part is to find M(e) in the...

oops.. I was meant to write m,n,o,p = 1 in teh original. hmm anyway.. for the actual question. the equivalence relation is "having the same prime factors(not neccessarily to the same power)". Prove the if part and the only if part separately. a).. trivial.. it's symmetric, it's...

Let me save turtle some work: Cut the infinite interval down to a finite interval: f(z) > 0 let E= f(z), there must exist c,d such that whenever x < c or when x > d, f(x) < E by the definition of the limits. consider the interval [c,d], there must be a maximum in the interval by the max...

^^ let me compress this: let m,n,o,p = 1 x|y implies abs(x) <= abs(y) y|x implies abs(y) <= abs(x) therefore y = +/- x similarly z= +/- y therefore z= +/- x so x | z^q, z| x^r for any q,r

alright alright.. here's a small encouragement: :D assuming you have the 2 case, let A= (x+y+z)/3 then A = [A + x + y + z]/4 = [(A+x)/2 + (y+z)/2]/2 >= [sqrt(Ax) + sqrt(yz)]/2 >= (Axyz)^(1/4) so A^(3/4) >= (xyz)^(1/4) A > (xyz)^(1/3) and the bad way to prove the 4 case...

Hmm.. Majority of the positions will stay the same.

87-92ish..

But the most likely application of L'hosptial's rule will be finding some limit for graph sketching.. and they don't care which method you used

Not as hard as you think, if it's soemthing between 0 and 1, it must be a half.

By inspection... in other words.. do it a few(~100) times then you should remember it

It isn't that difficult... I think it's asking you how a unit difference in polution (index of) will affect the wage level. Looks like economics

It's simple.. Whenever you get circle geometry questions, there's virtually only 2 things you could do.. and it's usually the first: angle bashing.

Learn the art of expressing advanced ideas in terms of those in the syllabus... and I heard some Uni lecturers and others mark the papers.. so they should understand anyway.

nope

(a,a) and (a,0) doesn't work..Suppose P is a point very close to (a,0)...

nope :D

I use to have 2 before 3 after school every fortnight. Going fast is good, gives you more time to revise