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

Question

Luke Alvarez · Accepted Answer

Not continuous

This function as is, is not continuous at #x_0=0# because it is not defined there. #D_f={x##in##RR##: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):}#

Paisley Johnson · Answer

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.

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)|=00}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.

Charles Arocha · Answer

undefined To prove that the function f(x) = x^2 sin(1/x) is continuous at x = 0, we need to show that the limit of f(x) as x approaches 0 exists and is equal to f(0).

First, let's evaluate the limit of f(x) as x approaches 0. We can rewrite the function as f(x) = x^2 * sin(1/x) = x * (x * sin(1/x)).

Now, we know that the limit of sin(1/x) as x approaches 0 is 0. This means that the term (x * sin(1/x)) approaches 0 as x approaches 0.

Next, we multiply this term by x, which also approaches 0 as x approaches 0. Therefore, the entire expression x * (x * sin(1/x)) approaches 0 as x approaches 0.

Since the limit of f(x) as x approaches 0 is 0, we can conclude that f(x) is continuous at x = 0.