How do you prove by definition that the function f(x)= x^2 sin (1/x) is continuous at x=0?

2 Answers
Dec 30, 2017

Not continuous

Explanation:

This function as is, is not continuous at x_0=0 because it is not defined there.

D_f={xinRR:x!=0} = RR* = (-oo,0)uu(0,+oo)

This function would be continuous for example,

f(x) = {(x^2sin(1/x)", "x!=0),(0" , "x=0):}

Dec 31, 2017

If you re-define the function as Jim suggested, then it would be continuous at x=0 and this can be proven as show below.

Explanation:

We must show that lim_{x->0}f(x)=f(0)=0.

Method 1 (use the epsilon/delta definition of a limit):

Let epsilon>0 be given (this represents an arbitrarily small distance that we'd like the function outputs to be to the limit 0 if x is sufficiently close to 0).

Choose delta=sqrt(epsilon)>0 (this represents the measure of "sufficiently close" and is chosen this way to make the algebra below work out nicely).

Suppose |x-0| < delta so that -sqrt(epsilon) < x < sqrt(epsilon). Then 0 leq x^{2} < (sqrt(epsilon))^{2}=epsilon and |x^{2}sin(1/x)| < epsilon (when x!=0), since -1 <= sin(1/x) <= 1 for all x!=0 (also note that |f(0)|=0<epsilon).

But this means that |f(x)-0|< epsilon for all x such that |x-0| < delta and we have shown that lim_{x->0}f(x)=f(0)=0, making f continuous at x=0.

Method 2 (use the Squeeze Theorem ):

The facts that -1 <= sin(1/x) <=1 for all x!=0 and f(0)=0 imply that -x^{2} <= f(x) <= x^{2} for all x. Since lim_{x->0}x^{2}=0 and lim_{x->0}(-x^{2})=0, it follows that lim_{x->0}f(x)=0=f(0) by the Squeeze Theorem.