Intuitively, we know that x^2 will go to 0 as x->0, and cos(pi/x) will be bounded by +-1, and so their product will go to 0 as x->0. We can be more formal about it using the definition of a limit, though.
We say that the limit as x->a of f(x) is L, denoted lim_(x->a)f(x)=L if for every epsilon>0 there exists a delta>0 such that 0<|x-a| < delta implies |f(x)-L| < epsilon.
We will use the above definition to show that lim_(x->0)x^2cos(pi/x)=0
Let epsilon>0 be arbitrary. Let delta=epsilon/(epsilon+1). Then if 0<|x-0| < delta, we have
|x^2cos(pi/x) - 0| = |x^2cos(pi/x)|
=|x^2|*|cos(pi/x)|
=|x|^2*|cos(pi/x)|
<= |x|^2*1" "(as |cos(theta)| <= 1 for all theta in RR)
=|x|^2
<= delta^2" "(as 0 < |x|=|x-0| < delta)
=(epsilon/(epsilon+1))^2
< epsilon/(epsilon+1)" "(as epsilon/(epsilon+1) < 1)
< epsilon
So, we have shown that for every epsilon > 0 there exists a delta > 0 such that 0 < |x-0| < delta implies |x^2cos(pi/x)-0| < epsilon, meaning lim_(x->0)x^2cos(pi/x) = 0.
∎
