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We safe $f$ is continuous at $x_0$ if for any $\epsilon >0$ there exists $\delta > 0$ such that whenever $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\epsilon$.

I'm reviewing for my final, and I have been coming up with functions and trying to show they are continuous. Here are the two I came up with: (1) $f(x)=x^2$ and (2) $f(x)=x^2-9$

To show (1) is continuous: $$|f(x)-f(2)|=|x^2-4|=|x+2||x-2|$$ Let $|x-2|<1$, then $1<x<3$, so $|x+2|<5$. $$|f(x)-f(2)|=|x^2-4|=|x+2||x-2|<5|x-2|$$

If we let $\delta=\min\{1,\epsilon/5\}$ we get what we wanted.

To show (2) is continuous at $x=2$ is a bit trickier I believe since it doesn't factor into pieces as evenly. This is where I'm stuck. Is there a way to rearrange the terms to get it to factor more nicely?

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This is sort of cheating, but you can always use the fact that the sum of two continuous functions is continuous. Since you just showed that $x^2$ is continuous, you then know that $x^2-9$ is also continuous. –  icurays1 Dec 7 '12 at 8:30
    
Did you get $x=2$ in 1. If so, why $x^2-9$? I think you should rewrite your solutions above again. –  B. S. Dec 7 '12 at 8:30
    
Also, $f(2)=4$ (not 9) in your first function. Shouldn't change things too much though. –  icurays1 Dec 7 '12 at 8:33
    
Fixed that. I have just been rewriting and modifying the problems in my notes. Errors have been propagating in the copying and studying. –  ortl Dec 7 '12 at 8:36

1 Answer 1

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For (1) it should be $|f(x)-f(2)|=|x^2-4|=|(x+2)(x-2)|.$

For (2) it does factor.
You have : $|f(x)-f(2)|=|x^2-9-2^2+9|=|x^2-4|=|(x+2)(x-2)|$.

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Ah clever. I was hastily allowing that to equal $-5$. Thanks for this. –  ortl Dec 7 '12 at 8:33

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