In Reply to: HW 8, question posted by Evan on March 01, 2003 at 20:48:24:
Evan,
>On proposition 12, you ask us to prove that f(x), and f(x_0) +
>f'(x-x_0), are e(x-x_0) close. But this does not seem possible, if
>for example you take f(x)=x, and then this does not hold for all
>e's. Am I missing something?
This is not exactly what Prop 12 says. It says that given any positive epsilon, there exists a positive delta such that f(x) and (f(x_0)+f'(x_0)*|x-x_0|) are eps|x-x_0|-close, given that x is in X and x is delta-close to x_0. It's basically saying that if you zoom in enough (small enough delta), that your function is approximately linear (i.e. the tangent line is epsilon-close to it) in that region, as long as the function is differentiable at x_0.
For example, if your function is f(x) = x, then the function is exactly linear so ANY delta will work: given any epsilon>0 and any x_0, you can choose any x in R and you know that f(x)-(f(x_0)+1*(x-x_0)) = x-x_0-x+x_0 = 0 < epsilon.
Does it make better sense this way? I think the key point is that x and x_0 have to be delta-close together.