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
