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Hanlin

General question about continuous functions

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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.

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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.

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(sorry about the rubbish ampersand, it's php applying specialchars, try and fix that asap)

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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

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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.

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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.

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I'm not 100% sure, but the proof goes something along the lines of using that subtraction formula in the definition of continuity.

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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.

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