How do you use the epsilon delta definition to find the limit of #x^2 cos(1/x)# as x approaches #0#?

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Answer 1 · 2016-12-08T17:01:45

#lim_(x->0) x^2cos(1/x) = 0#

The epsilon-delta formulation of the definition of limit states that:

#lim_(x->0) f(x) = 0#

if:

#AA epsilon >0, EE delta >0# such that for #|x| < delta, |f(x)| < epsilon#

So let's look at #f(x)# around #x=0# and note that:

#|x^2cos(1/x)|<=|x^2||cos(1/x)| <= |x^2|#

as #cosx# is bounded in the range #[-1,1]#.

So, given any #epsilon >0# take #delta < sqrt(epsilon#
Then for #x in (-delta, delta)#, of course #|x| < delta#, so that:

#|x^2cos(1/x)| <= |x^2| < delta^2 < epsilon#

which proves that:

#lim_(x->0) x^2cos(1/x) = 0#