Evaluate the limit? #lim_(x rarr 0) x^2(cos(1/x)-1) #

Question

Isabella Ackerman · Accepted Answer

I would use the squeeze theorem.

#-1<=cos(1/x)<=1# #" "# for all #x != 0# So , subtracting #1# from each part, #-2 <= cos(1/x)-1 <= 0# #" "# for all #x != 0# Note that #x^2 >0# #" "# for all #x != 0#, so we can multiply through without changing the inequalities. #-2x^2 <= x^2(cos(1/x)-1 ) <= 0# #" "# for all #x != 0# Since #lim_(xrarr0)(-2x^2) = 0 = lim_(xrarr0)0# the squeeze theorem tells us that #Lim_(xrarr0)x^2(cos(1/x)-1 ) = 0#

Aurora Alvarado · Answer

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

We have: # L = lim_(x rarr 0) x^2(cos(1/x)-1) # Let us perform a substitution: Let #u=1/x => x =1/u# Then as #x rarr 0 => u rarr oo # Substituting into the limit, we get: # L = lim_(u rarr oo) (1/u)^2(cos(u)-1) # # \ \ = lim_(u rarr oo) (cosu-1)/u^2 # # \ \ = lim_(u rarr oo) (cosu)/u^2-1/u^2 # # \ \ = lim_(u rarr oo) (cosu)/u^2- lim_(u rarr oo)1/u^2 # Consider the second limit: # lim_(u rarr oo)1/u^2 # This is a trivial limit as #u^2# increases without bound, thus #1/u^2 rarr 0# as #u rarr oo#, thus: # lim_(u rarr oo)1/u^2 = 0 # Consider now, the first limit: # lim_(u rarr oo) (cosu)/u^2 # Now, we have #-1 le cosu le 1 => -1/u^2 le cosu /u^2 le 1/u^2 #, thus: # lim_(u rarr oo)(-1/u^2) le lim_(u rarr oo)(cosu /u^2) le lim_(u rarr oo)(1/u^2) # And using the above result we just established that: # 0 le lim_(u rarr oo)(cosu /u^2) le 0 # So then by the sandwich, or squeeze, theorem we have: # lim_(u rarr oo)(cosu /u^2) = 0 # Thus: # L = 0 #

Emily Ammerman · Answer

undefined The limit of x^2(cos(1/x)-1) as x approaches 0 is 0.