Solved: "Solve the following limit:limx→π62sin⁡(x)−13tan⁡(x)−1"

SnuluddidaPalbx4

Answered question

2022-02-26

Solve the following limit:

lim_{x \to \frac{π}{6}} \frac{2 \sin (x) - 1}{\sqrt{3} \tan (x) - 1}

avalescogzw

Beginner2022-02-27Added 5 answers

F = \frac{2 \sin (x) - 1}{\sqrt{3} \tan (x) - 1} = \frac{2}{\sqrt{3}} \cdot \frac{\sin x - \sin \frac{π}{6}}{\tan x - \tan \frac{π}{6}} = \frac{2}{\sqrt{3}} \cdot \frac{2 \sin (\frac{x - \frac{π}{6}}{2}) \cos (\frac{x + \frac{π}{6}}{2})}{\tan x - \tan \frac{π}{6}}

(Applying

\sin C - \sin D = 2 \sin \frac{C - D}{2} \cos \frac{C + D}{2}

)
Now,

\tan x - \tan \frac{π}{6} = \frac{\sin (x - \frac{π}{6})}{\cos x \cos \frac{π}{6}}

So,

F = \frac{2}{\sqrt{3}} \cos x \cos \frac{π}{6} \frac{2 \sin (\frac{x - \frac{π}{6}}{2}) \cos (\frac{x + \frac{π}{6}}{2})}{2 \sin (\frac{x - \frac{π}{6}}{2}) \cos (\frac{x - \frac{π}{6}}{2})}

x \to \frac{π}{6}, \sin (\frac{x - \frac{π}{6}}{2}) \to 0 \Rightarrow \sin (\frac{x - \frac{π}{6}}{2}) \neq 0

So,

lim_{x \to \frac{π}{6}} F = lim_{x \to \frac{π}{6}} \frac{2}{\sqrt{3}} \cos x \cos \frac{π}{6} \frac{\cos (\frac{x + \frac{π}{6}}{2})}{\cos (\frac{x - \frac{π}{6}}{2})} = \frac{2}{\sqrt{}}

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