## The continuity of f

I have seen this theorem in a few books, but none of them give proofs, it says

if f(x) is a continuous function then lf(x)l is a continuous function. What is the proof of this because i don't really understand why this holds, thanks

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 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus The function $|~|:\mathbb{R}\rightarrow \mathbb{R}$ is continuous. Composition of continuous functions is continuous.
 It makes sense though. Consider y=-1/x : its continuous in the domain of x in (0,infinity) right? and its counterpart z=|-1/x| = 1/x is continuous in the same domain for x

## The continuity of f

Ahh, yeah i get that, but can you not prove it without using composition of two functions?

 You could probably prove it via the Weierstrass definition: http://en.wikipedia.org/wiki/Continuous_function but Im not a mathematician so you'll have to hope HallsOfIvy checks this thread.
 i might pm him, cheers
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus It is very easy to prove using the $\epsilon-\delta$ definition. Are you familiar with $\epsilon$and $\delta$ definitions?
 yeah, that's about the limit of my analysis knowledge, how would you use the ε,δ definition?

 Quote by macca1994 i might pm him, cheers
No need micro can handle it too. I think the epsilon / delta is the weierstrass definition.

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 Quote by macca1994 yeah, that's about the limit of my analysis knowledge, how would you use the ε,δ definition?
Take an $a\in \mathbb{R}$. You will need to prove

$$\forall \varepsilon >0: \exists \delta >0: \forall x: |x-a|<\delta~\Rightarrow ||f(x)|-|f(a)||<\varepsilon$$

You are given that f is continuous in a, thus:

$$\forall \varepsilon >0: \exists \delta >0: \forall x: |x-a|<\delta~\Rightarrow |f(x)-f(a)|<\varepsilon$$

So, take $\varepsilon>0$ arbitrary. Take $\delta>0$ as in the previous definition: so it holds that

$$\forall x: |x-a|<\delta~\Rightarrow |f(x)-f(a)|<\varepsilon$$

Take an x arbitrary such that $|x-a|<\delta$. Then we know that

$$||f(x)|-|f(a)||\leq |f(x)-f(a)|<\varepsilon$$

So we have verified that $||f(x)|-|f(a)||<\varepsilon$ and thus we have verified that $\varepsilon-\delta$ definition of continuity. Thus $|f|$ is continuous.

 ah that makes sense and is very obvious, do you need to show that limit of lf(x)l is in fact lf(a)l or is that just obvious?

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