We have:
Applying the Sandwich Theorem:
Hence:
Fernando Revilla
let f(x) = x^(2) sin (1/x^2) for 0<x< or = 1 and f(0)=0. prove that f is continuous on [0,1].
the answer said that since lx^(2) sin (1/x^2)l < or = l x^2 l, then f is continuous on 0 and hence continuous on [0,1]..but i do not see how this statement proves the question.
i thought to prove continuous, it would be assume that f is cont at c where c is between [0,1]
then
l x^(2) sin (1/x^2) - c^(2) sin (1/c^2)l < l x^2 - c^2l < 2 l x-cl
thus continuous?
We have:
Applying the Sandwich Theorem:
Hence:
Fernando Revilla
Implies that is continuous at .
I don't know why do you thought such a thing.i thought it just meant that the function is continuous in a small interval between 0? which might be smaller than 1?
Fernando Revilla
As you said before, "The only x where f(x) may not be continuous is x=0 (due to the zero in the denominator)." So in order to prove f is continuous on the interval [0, 1] it is only necessary to show that f is continuous at x= 0. And " is the definition of "continuous at x= 0".
Actually, showing [tex]\lim_{x\to a} f(x)= f(a)[/itex] only shows f is continuous at x= a, not on any interval at all. For example, the function f, defined by "f(x)= 1/n if x is a rational number, equal to m/n reduced to lowest terms, f(x)= 0 if x is irrational" is continuous only at x= 0, not on any interval around 0.i thought it just meant that the function is continuous in a small interval between 0? which might be smaller than 1?
But, again, as you said, continuity everywhere except x= 0 is obvious.
thanks!