How do you use the Squeeze Theorem to evaluate #lim x->0# of #x^2cos(1/(42x))#?

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Answer 1 · 2016-10-16T00:18:54

#lim_(x->0)x^2cos(1/(42x)) = 0#

As the range for the cosine function is #[-1, 1]#, we have that for all #x in RR#

#-1 <= cos(1/(42x)) <= 1#

#=> -x^2 <= x^2cos(1/(42x)) <= x^2#

Noting that

#lim_(x->0)(-x^2) = lim_(x->0)x^2 = 0#

we have, by the squeeze theorem,

#lim_(x->0)x^2cos(1/(42x)) = 0#