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so far i have said:

let [itex]\epsilon>0, \exists \delta>0 s.t. |x-x_0|< \delta[/itex]. now i need to show that [itex]|f(x)-f(x_0)|< \epsilon[/itex]. yes?

can't do the rest of it though.......

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- Thread starter latentcorpse
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- #1

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so far i have said:

let [itex]\epsilon>0, \exists \delta>0 s.t. |x-x_0|< \delta[/itex]. now i need to show that [itex]|f(x)-f(x_0)|< \epsilon[/itex]. yes?

can't do the rest of it though.......

- #2

Mark44

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could i just say x is continuous and 1-x^2 is continuous and non zero so f is continuous?

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Mark44

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- #5

HallsofIvy

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[tex]f(x)- f(x_0)= \frac{x}{1- x^2}- \frac{x_0}{1- x_0^2}[/tex]

[tex]= \frac{x(1-x_0^2)- x_0(1- x^2)}{(1- x^2)(1- x_0^2)}[/tex]

[tex]= \frac{x- xx_0^2- x_0- x^2x_0}{(1- x^2)(1- x_0^2)}[/tex]

[tex]= \frac{(x- x_0)- x_0x(x- x_0)}{(1- x^2)(1- x_0^2)}[/tex]

[tex]= (x- x_0)\frac{1- x_0x}{(1- x^2)(1- x_0^2)}[/tex]

Now you need to find a bound on

[tex]\frac{1- x_0x}{(1- x^2)(1- x_0^2)}[/tex]

You will, of course, want to stay away form x or [itex]x_0[/itex] being 1 or -1 so you might do this: Let a be the smaller of [itex]|1- x_0|[/itex] and [itex]|-1- x_0|[/tex] and require that [itex]|x- x_0|< a/2[/itex].

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