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Consider lim t → 0+ ( −2 sin(4t) sin(4t) + 2 t cos(4t) ) . using a table of values, the limiting value is

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Since this is not a rational function (with undetermined values in the denominator), but just a normal expression, we can just substitute t with the value toward which the function is approaching. Since this is a continuous function, it doesn't matter from which side it's approaching 0. In this case, let t=0, then sin(4t)=0, cos(4t)=1, then -2 sin^2(4t)+2tcos(4t)=0. 
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