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Find \mathrm{a} such that \mathrm{\lim _{x \rightarrow \frac{\pi}{6}}(\sqrt{3} \tan x)^{\frac{(a+1) \tan (3 x)}{\sin x}}=e^{\frac{-8 \sqrt{3}}{3}}}

Option: 1

\mathrm{a=2}


Option: 2

\mathrm{a=4}


Option: 3

\mathrm{a=-1}


Option: 4

\mathrm{a=0}


Answers (1)

best_answer

You need to find \mathrm{a} such that:
\mathrm{(a+1) \lim _{x \rightarrow \frac{\pi}{6}}\left(\frac{\sqrt{3} \tan x-1}{\sin x}\right) \tan 3 x=\frac{-8 \sqrt{3}}{3}}
or
\mathrm{2(a+1) \lim _{x \rightarrow \frac{\pi}{6}}(\sqrt{3} \tan x-1) \tan 3 x=\frac{-8 \sqrt{3}}{3}}

Now, let's evaluate the limit:

\mathrm{\lim _{x \rightarrow \frac{\pi}{6}}(\sqrt{3} \tan x-1) \tan 3 x =\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sqrt{3} \sin x-\cos x}{\cos x} \cdot \frac{\sin 3 x}{\cos 3 x} }
                                                      \mathrm{=\frac{4}{\sqrt{3}} \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \left(x-\frac{\pi}{6}\right)}{\cos 3 x}}
                                                      \mathrm{=\frac{4}{\sqrt{3}} \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \left(x-\frac{\pi}{6}\right)}{\cos 3 x}}
                                                      \mathrm{=\frac{4}{\sqrt{3}} \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \left(x-\frac{\pi}{6}\right)}{\cos x(1-2 \sin x)(1+2 \sin x)}}
                                                     \mathrm{=\frac{4}{3} \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \left(x-\frac{\pi}{6}\right)}{4 \cos \left(\frac{\pi}{12}+\frac{x}{2}\right) \sin \left(\frac{\pi}{12}-\frac{x}{2}\right)}}
                                                     \mathrm{ =\frac{2}{3 \sqrt{3}} \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \left(x-\frac{\pi}{6}\right)}{\sin \left(\frac{\pi}{12}-\frac{x}{2}\right)}}
                                                     \mathrm{ =-\frac{4}{3 \sqrt{3}} \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \left(x-\frac{\pi}{6}\right)}{x-\frac{\pi}{6}} \cdot \frac{\frac{\pi}{12}-\frac{x}{2}}{\sin \left(\frac{\pi}{12}-\frac{x}{2}\right)} }
                                                     \mathrm{ =-\frac{4}{3 \sqrt{3}} }
The final answer is \mathrm{ a=2 }.

Posted by

Kuldeep Maurya

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