Find $a$ such that $\lim_{x\to\frac{\pi}{6}}(\sqrt3 \tan x)^\frac{(a+1)\tan (3x)}{\sin x} = e^\frac{-8\sqrt3}{3}$

Question

What is the value of $a$ such that: $$\lim_{x\to\frac{\pi}{6}}\bigg(\sqrt3 \tan x\bigg)^\frac{(a+1)\tan (3x)}{\sin x} = e^\frac{-8\sqrt3}{3}$$

After using $\lim_{x\to 0}(1 + x)^\frac{1}{x}$, i have

$$e ^{\lim_{x\to\frac{\pi}{6}}\bigg(\frac{\sqrt3 \tan x - 1}{\sin x}\bigg) (a+1)\tan (3x)} = e^\frac{-8\sqrt3}{3}$$

Any hints for what to do next?

Answer 1

You need to find $a$ such that:

$$(a+1)\lim_{x\to\frac{\pi}{6}}\bigg(\frac{\sqrt3 \tan x - 1}{\sin x}\bigg)\tan 3x = \frac{-8\sqrt3}{3}$$

or

$$2(a+1)\lim_{x\to\frac{\pi}{6}} (\sqrt3 \tan x - 1) \tan 3x= \frac{-8\sqrt3}{3}$$

Now, let's evaluate the limit:

$$ \begin{aligned} \lim_{x\to\frac{\pi}{6}} (\sqrt3 \tan x - 1) \tan 3x &= \lim_{x\to\frac{\pi}{6}} \frac{\sqrt3 \sin x - \cos x}{\cos x}\cdot \frac{\sin 3x}{\cos 3x}\\ &=\frac{4}{\sqrt{3}}\lim_{x\to\frac{\pi}{6}} \frac{\sin\left(x-\frac{\pi}{6}\right)}{\cos 3x}\\ &=\frac{4}{\sqrt{3}}\lim_{x\to\frac{\pi}{6}} \frac{\sin\left(x-\frac{\pi}{6}\right)}{\cos 3x}\\ &= \frac{4}{\sqrt{3}}\lim_{x\to\frac{\pi}{6}}\frac{\sin\left(x-\frac{\pi}{6}\right)}{\cos x(1-2\sin x)(1+2\sin x)}\\ &= \frac{4}{3}\lim_{x\to\frac{\pi}{6}}\frac{\sin\left(x-\frac{\pi}{6}\right)}{4 \cos\left(\frac{\pi}{12} + \frac{x}{2}\right) \sin\left(\frac{\pi}{12} - \frac{x}{2}\right)}\\ &= \frac{2}{3\sqrt{3}}\lim_{x\to\frac{\pi}{6}}\frac{\sin\left(x-\frac{\pi}{6}\right)}{ \sin\left(\frac{\pi}{12} - \frac{x}{2}\right)}\\ &= -\frac{4}{3\sqrt{3}}\lim_{x\to\frac{\pi}{6}}\frac{\sin\left(x-\frac{\pi}{6}\right)}{x-\frac{\pi}{6}}\cdot \frac{\frac{\pi}{12} - \frac{x}{2}}{ \sin\left(\frac{\pi}{12} - \frac{x}{2}\right)}\\ &= -\frac{4}{3\sqrt{3}} \end{aligned} $$

The final answer is $a=2$.

Answer 2

If $$L=\lim_{x \rightarrow a} (f(x))^{g(x)} \rightarrow 1^{\infty}$$ then $$L=\exp[\lim_{x \rightarrow a} [g(x)(f(x)-1)].$$ So here $$L=\exp[\lim_{x\rightarrow \pi/6}[\frac{(a+1)\tan 3x}{\sin x}] [\sqrt{3} \tan x-1]]$$ $$L=\exp[\lim_{x\rightarrow \pi/6}[\frac{(a+1)}{\sin x}]\frac{ [\sqrt{3} \tan x-1]}{\cot 3x}]$$ Using L-Hospital's Rulr $$\lim_{x \rightarrow \pi/6} \frac{ [\sqrt{3} \tan x-1]}{\cot 3x}= \lim_{x \rightarrow \pi/6} \frac{\sqrt{3} \sec^2 a}{-3\csc^2 3x}=-\frac{4}{3 \sqrt{3}}$$ Then $$L=e^{-\frac{8(a+1)}{\sqrt{3}}} =e^{-\frac{8\sqrt{3}}{3}} \implies a=2.$$

Answer 3

If L'Hopital is allowed, the value of $a$ can be derived in relatively straightforward manner:

$$\begin{align} \lim_{x\to\pi/6}\left(\sqrt3\tan x \right)^{(a+1)\tan3x/\sin x}=e^{-8\sqrt3/3} &\implies\lim_{x\to\pi/6}{(a+1)\tan3x\over\sin x}\ln(\sqrt3\tan x)={-8\sqrt3\over3}\\ &\implies a+1={-8\sqrt3\over3}\lim_{x\to\pi/6}\left({\sin x\over\sin3x}{\cos3x\over\ln(\sqrt3\tan x)} \right)\\ &\implies a+1={-8\sqrt3\over3}\cdot{1/2\over1}\lim_{x\to\pi/6}\left(\cos3x\over\ln(\sqrt3\tan x)\right)\\ &\implies a+1={-4\sqrt3\over3}\lim_{x\to\pi/6}\left(-3\sin3x\over\sec^2x/\tan x \right)\\ &\implies a+1=4\sqrt3\lim_{x\to\pi/6}\left(\sin x\cos x\sin3x \right)\\ &\implies a+1=4\sqrt3\cdot{1\over2}\cdot{\sqrt3\over2}\cdot1=3\\ &\implies a=2 \end{align}$$

The key step is to check that $\lim_{x\to\pi6}\cos3x=\lim_{x\to\pi/6}\ln(\sqrt3\tan x)=0$, so that L'Hopital's Rule can be applied. If L'Hopital is not allowed (as the OP's tag indicates), $\lim_{x\to\pi/6}\cos3x/\ln(\sqrt3\tan x)$ can still be evaluated using the definition of the derivative:

$$\lim_{x\to\pi/6}{\cos3x\over\ln(\sqrt3\tan x)}={\displaystyle{\lim_{x\to\pi/6}{\cos3x-\cos(\pi/2)\over x-\pi/6}}\over\displaystyle{\lim_{x\to\pi/6}{\ln(\sqrt3\tan x)-\ln(\sqrt3\tan(\pi/6)\over x-\pi/6}}}={-3\sin(\pi/2)\over\sec^2(\pi/6)/\tan(\pi/6)}$$