Proving Continuity of f(x)=x^2sin(pi/x) at x=0

[zeroheero]

Hi, I have an assignment question that asks if f(x) = x^2sin(pi/x), prove that f(0) can be defined in such a way the f becomes continuous at x = 0.
Am I able to apply the squeeze theorem to show,
-1<sin(pi/x)<1
add x^2 to the inequality
-x\<x^2sin(pi/x)\<x^2. (\< us less than or equal to)
Lim as x approaches 0 from the left side, -x^2=0; and
Lim as x approaches 0 from the right side, x^2=0
if g(x) =-x^2 F(x) = x^2sin(pi/x). h(x)= x^2

Is this the best way tom get about this question?

 

[╔(σ_σ)╝]

Pretty much. It is the easiest way without invoking other theorems.