If you put the limiting value of the variable t=0 in to the function the see the function is become (0 / 0) form......
So, here you can use the l'hospital rule: that is you can take the derivative on numerator and denominator and after applying the firt order derivative you will have something like this: (2t) / ((2Cos t) * (Sin t)) = (2t) / Sin (2t).
now see we have the condition as : llim(t--->0) (2t) / Sin (2t).
See even if you the value t=0 in this fraction the form is (0/0). So, you have to apply the derivative on numerator and denominator also for the second time!!
So, if you apply the derivative you will get: lim(t--->0) (2) / (2Cos (2t)) ------(a)
Now we get rid of from the 0/0 form and now we can easily put the vallue t=0 in the function (a). and we have the limiting value as 1.
Hope the discussion will help you.best of luck.