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ferret93
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I'm working on a problem as part of exam revision, but I've run into a bit of trouble so far. The problem is;
Give an (ε,δ) proof that f(x) = 1/[itex]\sqrt{10 - x^2}[/itex] is continuous at x = -1
The attempt at a solution
So far what I've gotten is f(x) - f(-1) = 1/([itex]\sqrt{10 - x^2}[/itex]) - 1/3
= (3 - ([itex]\sqrt{10 - x^2}[/itex]))/(3[itex]\sqrt{10 - x^2}[/itex])
= ((x^2) - 1)/(3[itex]\sqrt{10 - x^2}[/itex])
Then from here I've gotten -2 < x < 0 → 10 - x^2 > 6
i'm lost from where to go from here though i just can't see a way though, any help will be greatly appreciated.
Give an (ε,δ) proof that f(x) = 1/[itex]\sqrt{10 - x^2}[/itex] is continuous at x = -1
The attempt at a solution
So far what I've gotten is f(x) - f(-1) = 1/([itex]\sqrt{10 - x^2}[/itex]) - 1/3
= (3 - ([itex]\sqrt{10 - x^2}[/itex]))/(3[itex]\sqrt{10 - x^2}[/itex])
= ((x^2) - 1)/(3[itex]\sqrt{10 - x^2}[/itex])
Then from here I've gotten -2 < x < 0 → 10 - x^2 > 6
i'm lost from where to go from here though i just can't see a way though, any help will be greatly appreciated.
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