Analysis - continuity of an algebraic fraction Watch

Swayum
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#1
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My analysis woes continue as I discover that f(x) = \frac{1}{\sqrt{x^2 - 4} - 1} is in fact not continuous at x = 2.

I was wondering how I could prove that (rather, where exactly the continuity proof would fail). My attempt is

Take any delta > 0 such that |x - 2| < \delta

Then |f(x) - f(2)| = |\frac{\sqrt{x^2 - 4}}{\sqrt{x^2 - 4} - 1}|

I want this to be less than epsilon eventually, so pretending like I've already found my epsilon, let's say that

|\frac{\sqrt{x^2 - 4}}{\sqrt{x^2 - 4} - 1}| < \epsilon

So then \sqrt{x^2 - 4} < \epsilon|\sqrt{x^2 - 4} - 1| \leq \epsilon(\sqrt{x^2 - 4} + 1) using the triangle inequality.

So a little rearranging gives \sqrt{x^2 - 4}(1 - \epsilon) \leq \epsilon

At this point I know I need to start getting delta involved. One way to do that is by noticing x^2 - 4 = (x-2)(x+2). So I start working on the left of my inequality above:

\sqrt{x^2 - 4}(1 - \epsilon) = \sqrt{x-2}\sqrt{x+2}(1 - \epsilon) > \sqrt{\delta} \sqrt{x+2}

Now obviously x + 2 > x - 2, so

\sqrt{x^2 - 4}(1 - \epsilon) > \sqrt{\delta} \sqrt{x+2} >  \sqrt{\delta} \sqrt{x - 2} > \sqrt{\delta}\sqrt{\delta} = \delta

So we can conclude that \epsilon \geq \sqrt{x^2 - 4}(1 - \epsilon) > \delta

So ultimately I find I just need epsilon > delta, i.e. take epsilon = delta + 1.

Is this correct?
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ghostwalker
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(Original post by Swayum)
Take any delta > 0 such that |x - 2| < \delta

Then |f(x) - f(2)| = |\frac{\sqrt{x^2 - 4}}{\sqrt{x^2 - 4} - 1}|
I'm v. rusty on analysis, but:

For x\in (-2,2), f(x) is not defined because of the square root. So there is no postive value for delta such that |f(x)-f(2)| is defined on the interval (2-\delta,2+\delta).

f(x) is actually continuous from the right at x=2.

Ignoring previous working, regarding "So ultimately I find I just need epsilon > delta, i.e. take epsilon = delta + 1". What if you take epsilon = 1/2, what's delta then?
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Swayum
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(Original post by ghostwalker)
I'm v. rusty on analysis, but:

For x\in (-2,2), f(x) is not defined because of the square root. So there is no postive value for delta such that |f(x)-f(2)| is defined in the interval (2-\delta,2+\delta).

f(x) is actually continuous from the right at x=2.
Oh ok I see.

Ignoring previous working, regarding "So ultimately I find I just need epsilon > delta, i.e. take epsilon = delta + 1". What if you take epsilon = 1/2, what's delta then?
Well no, I'm trying to prove that there exists some epsilon such that |f(x) - f(2)| < epsilon for all delta. So you give me delta, I find you epsilon. I'm not letting epsilon be arbitrary (I'm DISPROVING continuity, not proving it, so it suffices to find one epsilon).
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ghostwalker
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(Original post by Swayum)
Well no, I'm trying to prove that there exists some epsilon such that |f(x) - f(2)| < epsilon for all delta. So you give me delta, I find you epsilon. I'm not letting epsilon be arbitrary (I'm DISPROVING continuity, not proving it, so it suffices to find one epsilon).
Sorry, I thought you were doing a continuity proof and wanting to see where it failed.

If you wanted to disprove continuity, you'd need to find an epsilon, such that for all delta,.... So epsilon is chosen, a single value. But in that last part you've tied epsilon and delta in a 1-to-1 relationship, so there is only a single value for delta for a given epsilon.
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Swayum
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(Original post by ghostwalker)
Sorry, I thought you were doing a continuity proof and wanting to see where it failed.

If you wanted to disprove continuity, you'd need to find an epsilon, such that for all delta,.... So epsilon is chosen, a single value. But in that last part you've tied epsilon and delta in a 1-to-1 relationship, so there is only a single value for delta for a given epsilon.
So then how do you write down epsilon? I found above that I need epsilon > delta.

*Edit*

Ohhh I see what you're saying, you're saying that I've not really shown anything above because I've lower bounded |f(x) - f(c)| rather than upper bounding it.

So then, assuming |x - 2| < delta was valid, how would you proceed with the algebra?
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ghostwalker
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(Original post by Swayum)
...
I've read that post and started replying to it several times now, but it's just doing my head in trying to understand what you're asking.
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Mark13
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I think you're overcomplicating things: f(x) is definetly right-continuous at x=2, since

\displaystyle \lim_{x \rightarrow 2^+} \frac{1}{\sqrt{x^2-4}-1} = \frac{1}{\sqrt{\lim_{x \rightarrow 2^+}x^2-4}-1}=-1

by continuity of the various functions of which f is a composition etc.

The problem comes when we try to approach 2 from the left; \sqrt{x^2-4} is not real for -2&lt;x&lt;2, and so f (which presumably has been defined as a real function) is not defined for -2&lt;x&lt;2, so f does not have left-continuity at x=2, so is not continuous at x=2.
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