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#