For all #x != 0#, we have #-1 <= cos(sin(1/x)) <= 1#
Right Limit
For #0 < x < pi/2#, we have #tanx > 0#, so we can multiply the three parts of the inequality above by #tanx# without changing the inequalities.
For #0 < x < pi/2#, we get
#-tanx <= tanxcos(sin(1/x)) <= tanx#
Since #lim_(xrarr0^+)(-tanx )=0=lim_(xrarr0^+)(tanx )#, we have (by the right hand squeeze theorem)
#lim_(xrarr0^+)tanxcos(sin(1/x)) =0#
Left Limit
For #-pi/2 < x < 0#, we have #tanx < 0#, so when we multiply the three parts of the inequality by #tanx# we must change the inequalities.
For #-pi/2 < x < 0#, we get
#-tanx >= tanxcos(sin(1/x)) >= tanx#
Since #lim_(xrarr0^-)(-tanx )=0=lim_(xrarr0^-)(tanx )#, we have (by the right hand squeeze theorem)
#lim_(xrarr0^-)tanxcos(sin(1/x)) =0#
Twi-sided Limit
Because both the right and left limits at #0# are #0#, we conclude:
#lim_(xrarr0)tanxcos(sin(1/x)) =0#
