Prove that a function is continuous on an intervalby DeadOriginal Tags: continuous, function, interval, prove 

#1
Dec412, 02:59 PM

P: 274

1. The problem statement, all variables and given/known data
I have to prove that [itex]\sqrt{x}[/itex] is continuous on the interval [1,[itex]\infty[/itex]). 2. The attempt at a solution Throughout the school semester I believed that to show that a function is continuous everywhere all I need to do was show that [itex]\lim\limits_{h\rightarrow 0}f(x+h)f(x)=0[/itex] and I never thought much about it. It never really came up as a problem on any of the homeworks or exams so I never had much problem with it. I am now doing review problems for the final and I realized that f(x)=1/x is a clear counterexample to what I stated above. The limit as h approaches 0 of f(x+h)f(x) is 0 but 1/x is not continuous everywhere. Since this is the case, how do I show that a function is continuous everywhere within an interval? 



#2
Dec412, 03:26 PM

P: 1,309

To show it is continuous, you could go back to your first principles and use an epsilon delta argument.
Also given the interval, you can use the definition of the derivative, it's up to you. 



#3
Dec412, 03:55 PM

P: 274

Say I didn't know that [itex]\sqrt{x}[/itex] was differentiable. How would I use epsilon delta to prove that it was differentiable everywhere in that interval? The only epsilon delta proof for continuity that I know only proves that it is continuous at one point.
I read somewhere that I could choose my delta to not only depend on epsilon but also to depend on x. Is this correct? If this is, would it be right if I said Choose [itex]\delta=\epsilon(\sqrt{x}+\sqrt{a})[/itex]. Then [itex]xa<\delta\Rightarrowxa<\epsilon(\sqrt{x}+\sqrt{a})\Rightarrow xa\frac{1}{\sqrt{x}+\sqrt{a}}=\sqrt{x}\sqrt{a}<\epsilon[/itex]. 



#4
Dec412, 06:45 PM

HW Helper
Thanks
PF Gold
P: 7,176

Prove that a function is continuous on an interval\sqrt x \sqrt a = \frac {xa}{\sqrt x + \sqrt a}$$Given that you have ##x\ge 1## can you find an overestimate of the right side that doesn't involve ##x## in the denominator, then choose ##\delta## to make it work? 



#5
Dec412, 07:48 PM

P: 274

What if I then chose ##\delta=\min(\delta_{1},\epsilon(\sqrt{a1}+\sqrt{a}))##? Then my epsilon delta argument would be: Choose ##\delta=\min(\delta_{1},\epsilon(\sqrt{a1}+\sqrt{a}))##. Then ##xa<\delta\Rightarrow xa<\epsilon(\sqrt{a1}+\sqrt{a})\Rightarrow \sqrt{x}\sqrt{a}<\epsilon##. Also while we are talking about uniform continuity, if delta doesnt depend on ##a## for uniform continuity then what does delta depend on? Does it depend on x and y in that case? Like say I wanted to prove that ##\sqrt{x}## was uniformly continuous on [1,##\infty##) and I have ##xy<\delta##, can I then choose a ##\delta## that depends on ##x## and ##y## to show that ##\sqrt{x}\sqrt{y}<\epsilon##? 



#6
Dec412, 08:09 PM

HW Helper
Thanks
PF Gold
P: 7,176

\frac {xa} {\sqrt x + \sqrt a }\le \frac {xa} {1+\sqrt a}$$and use that to figure out ##\delta##? 



#7
Dec412, 08:24 PM

P: 274

##xa<\epsilon1+\sqrt{a}\Rightarrow xa\frac{1}{\sqrt{x}+\sqrt{a}}<\epsilon1+\sqrt{a}\frac{1}{1+\sqrt{a }}\Rightarrow \sqrt{x}\sqrt{a}<\epsilon##. Third times a charm? 



#8
Dec412, 08:30 PM

HW Helper
Thanks
PF Gold
P: 7,176





#9
Dec512, 09:19 AM

P: 274

##xa<\delta\Rightarrow xa<2\epsilon\Rightarrow xa\frac{1}{\sqrt{x}+\sqrt{a}}<2\epsilon\frac{1}{2}\Rightarrow \sqrt{x}\sqrt{a}<\epsilon##. Thank you! This has clarified a lot for me! 



#10
Dec512, 11:52 AM

HW Helper
Thanks
PF Gold
P: 7,176

$$ \sqrt x  \sqrt a = \frac{xa}{\sqrt x + \sqrt a}\le \frac{xa}{2}$$since ##x\ge 1## and ##a\ge 1##. (This is really the "exploratory argument" where you are figuring out what ##\delta## might work.) Suppose ##\epsilon>0##. Let ##\delta=2\epsilon##. Then if ##xa<\delta = 2\epsilon## we have$$ \sqrt x  \sqrt a <\frac{2\epsilon} 2=\epsilon$$ 


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