What does $\lim\limits_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$ evaluate to?

Question

What does $$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$ evaluate to? This very likely needs substitution.

Answer 1

We establish that the limit

$$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$

is of the indeterminate form $\dfrac00$; since numerator and denominator are differentiable, let us attempt De l'Hôpital's rule. It works:

$$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x} = \lim_{x\to\pi/6}\frac{-\sqrt3 \frac1{\cos^2 x}}{-6} = \frac{\sqrt3}6\cdot \frac43 = \frac{2\sqrt3}9$$

Answer 2

An idea with quite some trigonometry and algebra but without l'Hospital (not that there's something wrong with that, of course): substitute

$$y=x-\frac\pi6\iff x= y+\frac\pi6\;,\;\;\text{and}\;\;x\to\frac\pi6\implies y\to 0$$

so that our function now is

$$\begin{align*}&-\frac16\frac{1-\sqrt 3\tan\left(y+\frac\pi6\right)}{y}=\\ =&-\frac16\frac{1-\sqrt3\frac{\tan y+\frac1{\sqrt3}}{1-\frac1{\sqrt3}\tan y}}{y}\stackrel {\color{blue}{(*)}}=\\ =&-\frac16\frac{\rlap/\color{purple}1-\left(\sqrt3+\frac1{\sqrt3}\right)\tan y-\rlap /\color{purple}1}{y\left(1-\frac1{\sqrt3}\tan y\right)}\stackrel{\color{blue}{(**)}}=\\ =&\frac16\frac{4\tan y}{y\left(\sqrt3-\tan y\right)}=\end{align*}$$ $$=\frac16\frac{\tan y}y\frac4{\sqrt3-\tan y}\color{red}{\xrightarrow[y\to 0]{}}\frac16\cdot 1\cdot\frac4{\sqrt3}=\frac2{3\sqrt3}=\frac{2\sqrt3}9$$

$${}$$

$$\begin{align*}\;\;&(*)\;\;\frac{a-b\frac cd}{e}=\frac{ad-bc}{de}\\{}\\ \;\;&(**)\;\;\text{Factor out $\,\frac1{\sqrt3}\,$ in numerator and denominator and note minus sign in numerator}\end{align*}$$

Answer 3

By the L'Hôpital theorem $$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}=\lim_{x\to\pi/6}\frac{-\sqrt{3}(1+\tan^2 x)}{-6}=\frac{2\sqrt{3}}{9}$$

Answer 4

without L-hospital law $$\lim_{x\to\dfrac \pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$ $$\lim_{(x-\dfrac \pi6)\to 0}\frac{1-\sqrt{3}\dfrac {\sin x}{\cos x}}{6\left(\dfrac\pi 6-x\right)}$$ $$\lim_{(x-\dfrac \pi6)\to 0}\dfrac{1\cdot\cos x-\sqrt{3} \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$ $$\lim_{(x-\dfrac \pi6)\to 0}2\dfrac{\dfrac12\cdot\cos x-\dfrac {\sqrt{3}}{2} \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$ $$\lim_{(x-\dfrac \pi6)\to 0}2\dfrac {\sin \dfrac\pi6\cdot\cos x-\cos \dfrac \pi6 \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$ $$\lim_{(x-\dfrac \pi6)\to 0}\dfrac {\sin\left(\dfrac \pi6-x\right)}{3\left(\dfrac\pi 6-x\right)\cos x}$$

since $$\lim_{x\to a} \dfrac {\sin a}{a}=1$$ $$\lim_{(x-\dfrac \pi6)\to 0}\dfrac {1}{3\cos \dfrac \pi6}$$ $$\dfrac{2}{3\sqrt3}\implies \dfrac {2\sqrt3}{9}$$

Answer 5

$$\frac{1-\sqrt{3}\tan x}{\pi-6x}=\frac{\sqrt3}6\cdot\frac{\tan \frac\pi6-\tan x }{\frac\pi6-x}\left(\text{ as }\tan\frac\pi6=\frac1{\sqrt3}\right)$$ $$=\frac{\sqrt3}6\cdot\frac{\sin\left(\frac\pi6-x\right)}{\left(\frac\pi6-x\right)\cos x\cos \frac\pi6}$$

$$\lim_{x\to\frac\pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}=\frac{\sqrt3}{6\cos \frac\pi6}\cdot\lim_{x\to\frac\pi6}\frac{\sin\left(\frac\pi6-x\right)}{\left(\frac\pi6-x\right)}\frac1{\lim_{x\to\frac\pi6}\cos x}$$ $$=\frac{\sqrt3}{6\cos\frac\pi6}\cdot\lim_{y\to0}\frac{\sin y}y\cdot\frac1{\cos\frac\pi6}$$ (Putting $\frac\pi6-x=y$ in the first limit)

$$\lim_{x\to\frac\pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}= \frac{\sqrt3}{6\cos^2\frac\pi6}=\frac2{3\sqrt3}$$

---

$$\text{In fact, }\lim_{x\to a}\frac {f(x)-f(a)}{x-a}=f'(a)$$

$$\implies \lim_{x\to a}\left(\frac {\tan x-\tan a}{x-a}\right)=\left(\frac {d\tan x}{dx}\right)_{x=a}=\sec^2a$$

$$\implies \lim_{x\to \frac\pi6}\left(\frac {\tan x-\frac1{\sqrt3}}{x-\frac\pi6}\right)=\sec^2\frac\pi6=\frac43$$

Answer 6

I'm not sure why others used L'Hospital's rule but it's not necessary here. It works wonderfully but you (OP) may not yet be familiar with the technique. The limit is very close to being the definition of the derivative. We can rewrite it as

$$\lim_{x\rightarrow \frac{\pi}{6}}\frac{\sqrt{3}}{6}\frac{\tan x-\frac{1}{\sqrt{3}}}{x-\frac{\pi}{6}}.$$

The latter half of this is, in fact, just the definition of the derivative of $\tan x$ defined at $\frac{\pi}{6}$. Since we know what the derivative of it is, we can quickly write down the answer: $\frac{\sqrt{3}}{6}\left(\sec\left(\frac{\pi}{6}\right)\right)^2$.

Hopefully you can reduce this further.