@@drpeyam f(x0) is nonzero by assumption. We can do the rest by cases. Avraham did the positive already. For the negative case, we have f(x_0)=-|f(x_0)|. Then -|f(x0)|/2 < f(x) - (-|f(x0)|) < |f(x_0)|/2, or (-3/2)|f(x0)| < f(x) < (-1/2)|f(x0)|. Thus |f(x)| > (1/2)|f(x0)|. Though I think reverse triangle inequality is more elegant (and it extends to complex numbers!)
Ooh nice. I've only seen the "Proof by inscrutable choice of delta" version of this. Very nice how you motivated all the steps.
I showed 1/x is continuous (the boring sequence way) then used the fact that the composition of continuous functions is continuous
Prove your composition fact though, with epsilon delta
@@drpeyam lol of course !
Thanks for the proof!!
I think you can show more easily the inequality than having to use the reverse triangle inequality.
If |f(x)-f(x0)|
What if f(x0) is negative or 0?
@@drpeyam f(x0) is nonzero by assumption. We can do the rest by cases. Avraham did the positive already. For the negative case, we have f(x_0)=-|f(x_0)|. Then -|f(x0)|/2 < f(x) - (-|f(x0)|) < |f(x_0)|/2, or (-3/2)|f(x0)| < f(x) < (-1/2)|f(x0)|. Thus |f(x)| > (1/2)|f(x0)|. Though I think reverse triangle inequality is more elegant (and it extends to complex numbers!)
@@drpeyam I’m not sure what fails then
I was using the fact that |x-y|0 is equivalent to -z