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#.
∎
