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Question-93824




Question Number 93824 by student work last updated on 15/May/20
Answered by john santu last updated on 15/May/20
lim_(t→0)  lny = lim_(t→0)  tan (3t+(π/2))ln(tan (((3t)/2)+(π/4)))  = lim_(t→0)  ((ln(tan (((3t)/2)+(π/4))))/(cot (3t+(π/2))))  = lim_(t→0)  ((ln(tan (((3t)/2)+(π/4))))/(−tan 3t))  now easy to solve
$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{lny}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{tan}\:\left(\mathrm{3t}+\frac{\pi}{\mathrm{2}}\right)\mathrm{ln}\left(\mathrm{tan}\:\left(\frac{\mathrm{3t}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\mathrm{tan}\:\left(\frac{\mathrm{3t}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)}{\mathrm{cot}\:\left(\mathrm{3t}+\frac{\pi}{\mathrm{2}}\right)} \\ $$$$=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\mathrm{tan}\:\left(\frac{\mathrm{3t}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)}{−\mathrm{tan}\:\mathrm{3t}} \\ $$$$\mathrm{now}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\: \\ $$

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