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Find \lim _{t \rightarrow 0} \frac{\tan t \sec 2 t}{3 t}
 

Option: 1

0


Option: 2

\frac{2}{3}


Option: 3

\frac{1}{3}


Option: 4

\frac{4}{5}


Answers (1)

best_answer

Solution From the definition of tan \mathrm{t} and \mathrm{\sec 2 t}, we have

\mathrm{\lim _{t \rightarrow 0} \frac{\tan t \sec 2 t}{3 t} =\lim _{t \rightarrow 0} \frac{1}{3} \cdot \frac{1}{t} \cdot \frac{\sin t}{\cos t} \cdot \frac{1}{\cos 2 t} }

\mathrm{ =\frac{1}{3} \lim _{t \rightarrow 0} \frac{\sin t}{t} \cdot \frac{1}{\cos t} \cdot \frac{1}{\cos 2 t} }

\mathrm{ =\frac{1}{3}(1)(1)(1)=\frac{1}{3} . } Eq. (1) and Example I Ib in Section 2.2
More trigonometric limits:
We have already seen \mathrm{ \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1 }. Similarly we can say

\mathrm{ \lim _{\theta \rightarrow 0} \frac{\theta}{\sin \theta}=1 ; \lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1 ; \lim _{\theta \rightarrow 0} \frac{\theta}{\tan \theta}=1 ; \lim _{\theta \rightarrow 0} \frac{\sin ^{-1} \theta}{\theta}=1 ; }

\mathrm{ \lim _{\theta \rightarrow 0} \frac{\theta}{\sin ^{-1} \theta}=1 ; \lim _{\theta \rightarrow 0} \frac{\tan ^{-1} \theta}{\theta}=1 ; \lim _{\theta \rightarrow 0} \frac{\theta}{\tan ^{-1} \theta}=1 }

Hence option 3 is correct



 

Posted by

sudhir kumar

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