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Hanlin

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About Hanlin

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  1. 2004-06-30 Hello. My name is Hans Lindroth. I have a question about continuity. How do you verify/proove that, for instance, arcsine(x) is continuous? I mean, when I apply the definition of continuity on it I get: 0<=|arcsine(x) - arcsine(a)|<e (e>0) (Here "<=" means "smaller than or equal to"). I'm stuck here however since I don't know of any method to convert "arcsine(x) - arcsine(a)" into something I can modify. Can anybody help me out? If you can, please give me a hint. Thanks Hans Lindroth.
  2. 2004-06-30 Hi. My name is Hans Lindroth. I study maths at the university for the first time and was hoping someone could answer me this. If I'd like to find the limit of, say, xarctan(1/x) lim x->0 How would I do it? I mean, does the function arctan(1/x) go to infinity as (1/x) does (when x->0) or does it just oscillate like arcsine or arccos? If anybody knows, please tell me. Thanks Hans Lindroth.
  3. 2004-06-30 Hello, Dave. I finally came up with a solution to the problem, thanks to you If you use the definition for continuity (like you said), combined with the subtraction theorum for sine and the fact that sin(u)<=|u| and cos(u)<=1 then what you get is this: (here "<=" means "smaller than or equal to"). 0<=|sin(x) - sin(a)| = (subtraction formula for sine) = 2|sin(x-a)/2|*|cos(x+a)/2|<= (cos(u)<=1, sin(u)<=|u|) <= 2|(x-a)/2| <=|x-a| This means that the function |sin(x) - sin(a)| is "trapped" between 0 and the function |x - a|, which goes to 0 as x - >a. Hence: lim |sin(x) - sin(a)| = 0 x->a which is the same as if: lim sin(x) = sin(a) x->a For all x. Q.E.D Thanks for the hint Hans Lindroth.
  4. Thank you, that really helped;) Now here's a harder problem of the same kind. It looks really nasty. It goes like this: "Show that sinx is a continuous function by using the subtraction formula sin(u)-sin(v) = 2sin((u-v)/2)*cos((u+v)/2) and the inequality |sinx|<=|x| (here <= means 'smaller than or equal to') and basic postulates concerning limits of functions". I'll have a go myself. But I could probably need some help. Thanks again Hans Lindroth.
  5. Thanks for helping me Dave. I was just wondering what you mean by sequential continuity. Is it a method? Please clarify. Thanks again Hans Lindroth
  6. Hello. My name is Hans Lindroth. I'm studying maths at the university for the first time. This time I have a question of more general type. It goes like this: "show, by using the basic postulates concerning limits of functions, that |x| is a continuous function." How should I approach this problem? If anybody can give me a hint then please help me. Thanks Hans Lindroth.
  7. Newbie Group: Members Posts: 5 Member No.: 295 Joined: 26-June 04 2004-06-27 Hello there. My name is Hans Lindroth. I've just started to study maths at the university. Now I've encountered a problem about limits. It goes like this: "In a circle with radius 'R' a chord is drawn, the length of which is 'L'. Let 'H' be the distance between the center of the chord and the center of the smaller arc being cut off the circle by the chord. Decide the limit: lim H/(L^2)". L->0 I've tried by finding an expression for "H" in terms of "L". This doesn't seem to lead me right though. Besides, the answer (which is "1/8(R^2)") is given in terms of 'R' (which means I must have lost myself). If you know something I don't, please help me. Thanks Hans Lindroth.
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