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leehuan's All-Levels-Of-Maths SOS thread (2 Viewers)

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leehuan

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Try using L'Hôpital's rule or a substitution.
Can't use the former cause haven't actually been taught it yet (or I would've). Will try a substitution later I guess, thanks.
 

Drongoski

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cancelling the common factors, as you were taught to do in your HSC - but were not told why you can do this. You would now know the reason, from learning the delta-epsilon definition of a limit.

To use L'Hopital's rule, you simply differentiate the numerator and the denominator wrt 'x', and substitute x=64 in the resultant expression; if still indefinite, you can reapply the rule until no longer indefinite, like the original expression would have given you a so-called indefinite 0/0.
 
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leehuan

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I know how to use L'Hopital's; I've used it before. I just haven't been taught it in my course yet to use it in my homework.

But I see. I was looking for something to cancel out so I guess the fractional powers threw me off.
 

leehuan

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I have no idea what's happening, so I'm just gonna say "Geometric Series"
It has absolutely nothing to do with geometric series; this is all related to nothing more than limits.
 

InteGrand

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The answer to b) is a correct one.

The answer in c) is a correct one too. To see this, let eps > 0, N = 1/eps^2, and suppose x > N.

Then x > 1/eps^2, which implies sqrt(x) > 1/eps. This implies that sqrt(x) > (1/eps) – 1, and this is equivalent to 1/(1 + sqrt(x)) < eps, i.e. |1/(1+sqrt(x)) – 0| < eps. Hence if x > N, then |f(x) – L| < eps, so the N is a correct answer.
 

InteGrand

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An easier way to see that N works is that for any 0 < eps < 1, we have 1/eps > (1/eps – 1) > 0, so (1/eps^2) > (1/eps – 1)^2, that is, N > M. Since we know that M works (by 'works', I mean that x being greater than it will imply that |f(x) – L| < eps), taking any larger value (e.g. N) will also work. This is because if x > N, then x > M (as N > M), implying that |f(x) – L| < eps, so N works too.
 

leehuan

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Ohhh should've trusted my very first instinct then. Thanks
 

leehuan

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Lost in my algebra today big time.



textbf - what I'm stuck on basically
 

leehuan

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Nothing to do with instinct. If you had said 'Yes', you'd still have to justify your answer.



Not sure if you understood what I meant.

My first instinct processed almost the ENTIRE method. But for some reason I had a contradiction at one point in my mind which halted me. And for some reason I went with that.
 

Drongoski

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Not sure if you understood what I meant.

My first instinct processed almost the ENTIRE method. But for some reason I had a contradiction at one point in my mind which halted me. And for some reason I went with that.
Ok
 

Green Yoda

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Question:
Determine the centre and radius of the circle.
x^2+y^2-18x+20y+60=0
 
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