Hanlin Posted June 27, 2004 Share Posted June 27, 2004 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. Link to comment Share on other sites More sharing options...

Dave Posted June 27, 2004 Share Posted June 27, 2004 I think I told you in your PM, but just for clarity's sake: [math]|x| = \begin{cases}x,& x\geq 0\\ -x,& x \leq 0 \\ \end{cases}[/math]. x and -x are very easy to prove continuous (sequential continuity is probably the easiest way) and if you show the limit exists at 0 (i.e. [math]\lim_{x\to 0} (x) = \lim_{x\to 0} (-x)[/math]) then you've got your proof. Link to comment Share on other sites More sharing options...

Dave Posted June 27, 2004 Share Posted June 27, 2004 (sorry about the rubbish ampersand, it's php applying specialchars, try and fix that asap) Link to comment Share on other sites More sharing options...

Hanlin Posted June 27, 2004 Author Share Posted June 27, 2004 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 Link to comment Share on other sites More sharing options...

Dave Posted June 27, 2004 Share Posted June 27, 2004 Nope. Basically, this is the definition: A function [math]f(x):\mathbb{R}\to\mathbb{R}[/math] is sequentially continuous at a point c if and only if for any sequence [math](x_n) \to c, f(x_n) \to f©[/math] for some [math]x\in\mathbb{R}[/math]. A consequence of this is that every continuous function is sequentially continuous and vice versa. So for f(x) = x, take some sequence [math]a_n \to c[/math]. Then [math]f(a_n) = a_n[/math], so [math]\lim_{n\to\infty} f(a_n) = \lim_{n\to\infty}a_n = c = f©[/math] as required. This effectively proves that f(x) is continous for all of the reals - similar proof for f(x) = -x. Link to comment Share on other sites More sharing options...

Hanlin Posted June 27, 2004 Author Share Posted June 27, 2004 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. Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Share Posted June 28, 2004 I'm not 100% sure, but the proof goes something along the lines of using that subtraction formula in the definition of continuity. Link to comment Share on other sites More sharing options...

Hanlin Posted June 30, 2004 Author Share Posted June 30, 2004 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. Link to comment Share on other sites More sharing options...

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