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Category: Limits

Question-151538

Question Number 151538 by DELETED last updated on 21/Aug/21 Answered by DELETED last updated on 21/Aug/21 3).limt[(sin2t)3t].t6=..?=limt0[(sin2t)3t].16t$$\:\:\:\:\:\:=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\left(\frac{\mathrm{sin}\:\mathrm{2t}}{\mathrm{6t}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\right] \