# Evaluating $\lim_{x \to \frac{\pi}{6}}{(1-2\sin(x))}^{\tan(\frac{\pi}{6}-x)}$

I'm in a little struggle with this limit, can anyone help me, please?

$$\lim_{x \to \frac{\pi}{6}}{(1-2\sin(x))}^{\tan(\frac{\pi}{6}-x)}$$

I tried to use the logarithm to then use L'Hospital's rule but I got stuck here: $$\ln(L)=\lim_{x \to \frac{\pi}{6}}{[\tan(\frac{\pi}{6}-x)\ln(1-2\sin(x))]}$$

Thank you!

• First bring it into the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and then apply the mentioned theorem. – MathAnimal May 20 at 10:05
• FYI L'Hopital Rule is applied when there is fraction. See this. – AryanSonwatikar May 20 at 10:06
• I did it but it leads to that logarithm/( 1/0 ) and 1/0 doesn't exist – Cookie Dog May 20 at 10:06
• write it as $$\frac{\ln(1-2 \sin x)}{\cot \left(\frac{\pi}{6}-x\right) }$$ and use the fact that $\cot \left(\frac{\pi}{6}-x\right) \to \infty$ and $\ln(\ldots) \to ????$as $x \to \frac{\pi}{6}$. – Anurag A May 20 at 10:09
• Are you sure the limit is meant to be $2$-sided? If $x$ is slightly larger than $\pi/6$, $1-2\sin x<0$, so $(1-2\sin x)^{\tan(\pi/6-x)}$ isn't real. – J.G. May 20 at 13:45

Let $$f(x) = (1-2\sin x)^{\tan(\frac{\pi}{6}-x)}$$, then $$f(x) = e^{g(x)}$$ with $$g(x) = \tan(\frac{\pi}{6}-x) \log (1-2\sin x)$$.

\begin{align} \lim\limits_{x \to \frac{\pi}{6}^- } g(x) &= \lim\limits_{x \to \frac{\pi}{6}^- } \frac{\tan\left(\frac{\pi}{6}-x\right)}{\frac{\pi}{6}-x} \left(\frac{\pi}{6}-x\right)\log \left(1-2\sin x\right) \\ &\overset{(1)}{=} \lim\limits_{x \to \frac{\pi}{6}^- } \left(\frac{\pi}{6}-x\right)\log \left(1-2\sin x\right) \\ &=\lim\limits_{x \to \frac{\pi}{6}^- } \frac{ \log (1-2\sin x)}{\frac{1}{\frac{\pi}{6}-x}} \\ &\overset{\mathrm{H}}{=} \lim\limits_{x \to \frac{\pi}{6}^-} (-2\cos x)\frac{\left(\frac{\pi}{6}-x\right)^2}{1-2\sin x} \\ &= -\sqrt{3}\lim\limits_{x \to \frac{\pi}{6}^-} \frac{\left(\frac{\pi}{6}-x\right)^2}{1-2\sin x} \\ &\overset{\mathrm{H}}{=} -\sqrt{3}\lim\limits_{x \to \frac{\pi}{6}^-}\frac{-2\left(\frac{\pi}{6}-x\right)}{-2\cos x } \\ &= 0 \end{align} where in $$(1)$$ I have used $$\lim_{y\to0} \frac{\tan y}{y} = 1$$ and $$H$$ denotes the usage of L'Hôpital's rule.

Hence, we conclude that

$$\lim\limits_{x \to \frac{\pi}{6}^-} f(x) = e^0 = 1.$$

• Thanks but why did you wrote $x \to \frac{\pi}{6}^{-}$? My limit is a two-sided one. – Cookie Dog May 20 at 13:06
• @user3669039 I've edited the formatting of your algebra for readability. It is generally easier for people to follow equations separated by lines instead of written sequentially on the same line. – Jam May 20 at 13:43
• @Jam Thank you very much – user3669039 May 20 at 13:44
• @CookieDog The limit cannot be two-sided: we must guarantee $1-2\sin(x)>0$ in order for your function to be defined. – user3669039 May 20 at 13:51

In the "$$\log$$" expression, expand $$\sin$$ around $$x_0=\frac{\pi}{6}$$ using Taylor series, up to the second term, get $$\sin x \approx \frac{1}{2} + \frac{\sqrt{3}}{2}\left(x-\frac{\pi}{6}\right)$$ so the expression $$\log(1-2 \sin x)$$ becomes $$\log\left(\frac{\sqrt{3}}{2} \left(\frac{\pi}{6} - x\right)\right) = \log \frac{\sqrt{3}}{2} + \log \left(\frac{\pi}{6} - x\right)$$. Now set $$t=\frac{\pi}{6} - x$$, rewrite $$-\tan (-t_ = -\frac{\sin t}{\cos t}$$ and expand $$\sin t \sim t$$ for $$t \to 0^+$$. This additional condition of convergence from the right allows rewriting the limit as

$$\lim_{t \to 0^{+}} t \log t$$

Now you can rewrite $$t \log t = \frac{\log t }{\frac{1}{t}}$$, and note that $$\frac{1}{t} \to \infty$$ and $$\log t \to -\infty$$. Set $$\log t =v, \frac{1}{t} = e^{-v}$$ for $$v \to \infty$$ and obviously $$\lim_{v \to \infty}\frac{v }{e^v} = 0$$ All other terms converge to constants and are easy to compute. Keep in mind also that the original expression is $$\varphi = e^{\log \varphi}$$, so don't forget to take the exponent.

Result: no L'Hopital Rule used, only Taylor Series Expansion!

• There is a problem with your Taylor expansion, it should be $\frac{1}{2}-\frac{\sqrt{3}}{2}(\frac{\pi}{6}-x)$ – user3669039 May 20 at 12:39
• thanks @user3669039, fixed – Alex May 20 at 15:47

Your job might be simpler if you substitute $$\pi/6-x=2t$$. Then $$1-2\sin x=1-2\sin(\pi/6-2t)=1-\cos 2t+\sqrt{3}\sin 2t=2\sin t(\sin t+\sqrt{3}\cos t)$$ Note that in order that the limit makes sense you need $$\sin x<1/2$$, so $$0 (the lower bound is mostly irrelevant, though), hence $$t>0$$.

How does this help? You get to evaluate the limit for $$t\to0$$ of $$\tan2t\bigl(\log(\sin t)+\log(2\sin t+2\sqrt{3}\cos t)\bigr)$$ The part $$\tan2t\log(2\sin t+2\sqrt{3}\cos t)$$ poses no problem: its limit is $$0$$. Then you need to compute the limit of $$\frac{2\cos t}{\cos2t}\sin t\log\sin t$$ The fraction part has limit $$2$$. The part $$\sin t\log\sin t$$ has limit $$0$$, as it's easy to show with l'Hôpital or other methods.

Hence the limit is $$0$$. Therefore your original limit is $$e^0=1$$.