1. ## continuity

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?

2. We have:

$\displaystyle 0\leq |f(x)|\leq x^2\quad (\forall x\in [0,1])$

Applying the Sandwich Theorem:

$\displaystyle \displaystyle\lim_{x \to 0}|f(x)|=0$

Hence:

$\displaystyle \displaystyle\lim_{x \to 0}f(x)=0=f(0)$

Fernando Revilla

3. Originally Posted by alexandrabel90
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?
The only x where f(x) may not be continuous is x=0 (due to the zero in the denominator).

4. why does showing that the limit of the function = 0 imply that the function on continuous from 0 to 1?

i thought it just meant that the function is continuous in a small interval between 0? which might be smaller than 1?

thanks!

5. Originally Posted by alexandrabel90
why does showing that the limit of the function = 0 imply that the function on continuous from 0 to 1?
Implies that is continuous at $\displaystyle 0$.

i thought it just meant that the function is continuous in a small interval between 0? which might be smaller than 1?
I don't know why do you thought such a thing.

Fernando Revilla

6. Originally Posted by alexandrabel90
why does showing that the limit of the function = 0 imply that the function on continuous from 0 to 1?
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 "$\displaystyle \lim_{x\to 0}f(x)= f(0)$ is the definition of "continuous at x= 0".

i thought it just meant that the function is continuous in a small interval between 0? which might be smaller than 1?
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.

But, again, as you said, continuity everywhere except x= 0 is obvious.

thanks!