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Showing a fucntion is continuous on an interval

  • Thread starter k3k3
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  • #1
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Homework Statement


Use the definition of continuity to prove that the function f defined by f(x)=x^(1/2) is continuous at every nonnative number.


Homework Equations


Continuity in this text is defined as

Let I be an interval, let f:I→ℝ, and let c be in I. The function f is continuous at c if for each ε>0 there exists a δ>0 such that |f(x)-f(c)|<ε for all x in I that satisfy |x-c|<δ.


The Attempt at a Solution



For x=0 is easy. I've shown that.

For x>0,

Let ε > 0
Let I=[0,∞)

Find δ > 0 such that |f(x)-f(c)| < ε for all x in I that satisfy |x-c|<δ

I know |x^(1/2)-c^(1/2)| ≤ |x-c| intuitively.

I've tried saying that since |x^(1/2)| ≤ |x| and |-(c^(1/2)| ≤ |-c| => |x^(1/2)| + |c^(1/2)| ≤ |x|+|-c| => |x^(1/2)+c^(1/2)|≤ |x| + |c|

This is where I am stuck. I am not sure how to show this properly.
 

Answers and Replies

  • #2
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|x-c|<δ corresponds to (c-δ) < x < (c+δ). Choose either and show that |f(x)-f(c)| < ε, where ε goes to 0 if δ goes to 0.
I think you will have to use the approximation (1+x)n = 1 + nx when x << 1.
 
  • #3
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Well, first off, it isn't true that |x^(1/2)| ≤ |x|. What if x=0.25? Then x^(1/2)=0.5, so the inequality fails.

Let's look at [itex]|f(x)-f(c)|=|\sqrt{x}-\sqrt{c}|[/itex]. If you multiply the top and the bottom of the expression in modulus by [itex]\sqrt{x}+\sqrt{c}[/itex], then you get
[tex]|f(x)-f(c)|=\left|\frac{x-c}{\sqrt{x}+\sqrt{c}}\right|=\frac{|x-c|}{\sqrt{x}+\sqrt{c}}.[/tex]
Supposing [itex]|x-c|<\delta[/itex] for some delta that is yet to be determined, you have
[tex]|f(x)-f(c)|<\frac{\delta}{\sqrt{x}+\sqrt{c}}.[/tex]
We're nearly there... we just need to choose delta so that this expression is less than epsilon. However, we still have x in our denominator, and delta shouldn't depend on x, only c. Can you think of a way of manipulating this so that delta is a bit more... obvious to choose?

Don't look at the spoiler below until you've tried at least 10 minutes to figure this out, otherwise you won't learn. Real Analysis is considered the most difficult 2nd-year math course at my university, and if yours is anything like mine, then you will need to practice.

Hint: [itex]\delta=\epsilon\sqrt{c}[/itex] will work...
 
  • #4
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|x-c|<δ corresponds to (c-δ) < x < (c+δ). Choose either and show that |f(x)-f(c)| < ε, where ε goes to 0 if δ goes to 0.
I think you will have to use the approximation (1+x)n = 1 + nx when x << 1.
The OP has said that their definition of continuity is the classic epsilon-delta definition. If they haven't seen the equivalent (rigorous) definition using limits, then I think it's inappropriate to suggest this approach.
 
  • #5
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The OP has said that their definition of continuity is the classic epsilon-delta definition. If they haven't seen the equivalent (rigorous) definition using limits, then I think it's inappropriate to suggest this approach.
Fair enough.
 
  • #6
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Thanks guys. I will work on it some more before looking at the spoiler.
 

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