Understanding Continuity and Intervals for Limits in Functions

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The discussion focuses on understanding the continuity of the function f(x) = 1/x^3 and how limits relate to it. A function is continuous at a point if the limit as h approaches 0 of [f(x+h) - f(x)] equals zero. The continuity of f is dependent on the continuity of the function 1/x, which is undefined at x = 0, thus making f discontinuous there. The participants clarify that a function cannot be continuous where it is undefined, emphasizing the importance of identifying the domain. Overall, the conversation aims to demystify limits and continuity in the context of specific functions.
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i would like someone to clear my doubts by explaining how it actually works..limits is one of the chapter i fear about it cause I am quite blur with it..so i do need someone help me;v this que..given f(x)=1/x^3 what r the intervals for function continuous ?how to solve it?thanks :)
 
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f is continuous when
lim_{h->0} [f(x+h)-f(x)]=0
also composition of continuous funtions are continuous
x^3 is everywhere continuous
where is 1/x continuous?
 
lurflurf said:
f is continuous when
lim_{h->0} [f(x+h)-f(x)]=0
also composition of continuous funtions are continuous
x^3 is everywhere continuous
where is 1/x continuous?

u meant composite of continuous function could also be a continuous for x^3?bt the limit is not given?!so what will be the interval for the function?how to differentiate between them and i still don't really get your point;i'm so sorry!
 
let g(x)=x^3
write f(x)=1/x^3
as
f(g(x))=g(1/x)=(1/x)^3
g(x) is continuous for all real numbers (show this)
so f is continuous at a if and only if 1/x is continuous at a.
Where is 1/x continous?

Start by giving the domain of 1/x
A function cannot be continuous where it is undefined.
 
lurflurf said:
let g(x)=x^3
write f(x)=1/x^3
as
f(g(x))=g(1/x)=(1/x)^3
g(x) is continuous for all real numbers (show this)
so f is continuous at a if and only if 1/x is continuous at a.
Where is 1/x continous?

Start by giving the domain of 1/x
A function cannot be continuous where it is undefined.

hmm,then i think i get your point..anyway thanks alot...:smile:
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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