Your classmate requests you to help him/her in understanding the solutions of the
lim
t→0
(1−cos3t)/
(3t)
= 0, via writing. How do you apply your knowledge based on what you have
learned from this module in doing the task?
Ans:-
limt→01−Cos3t3t⇒limt→01−(1−2Sin23t)3t⇒limt→02Sin23t3t\lim_{t \to 0} \dfrac{1-Cos3t}{3t}\\ \Rightarrow \lim_{t \to 0} \dfrac{1-(1-2Sin^23t)}{3t}\\ \Rightarrow\lim_{t \to 0} \dfrac{2Sin^23t}{3t}\\ \\limt→03t1−Cos3t⇒limt→03t1−(1−2Sin23t)⇒limt→03t2Sin23t
⇒limx→02×Sin23t×3t(3t)2⇒limt→06tlimt→0Sin23t(3t)2\Rightarrow \lim_{x \to 0} 2\times\dfrac{Sin^23t\times3t}{(3t)^2}\\ \Rightarrow\lim_{t \to 0}6t\lim_{t \to 0} \dfrac{Sin^23t}{(3t)^2}⇒limx→02×(3t)2Sin23t×3t⇒limt→06tlimt→0(3t)2Sin23t ∵limt→0Sintt=1\because \lim_{t \to 0} \dfrac{Sint}{t}=1∵limt→0tSint=1
⇒limt→06t×1\Rightarrow \lim_{t \to 0}6t \times1⇒limt→06t×1 ∵limt→0Sin23t(3t)2=1\because \lim_{t \to 0} \dfrac{Sin^23t}{(3t)^2}=1∵limt→0(3t)2Sin23t=1
⇒0\Rightarrow 0⇒0 .....Ans
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