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

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

I tried to use the logarithm to then use L’Hospital’s rule but I got stuck here:

$ln(L)=lim_{x to frac{pi}{6}}{[tan(frac{pi}{6}-x)ln(1-2sin(x))]}$

Thank you!

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

$$begin{align}

limlimits_{x to frac{pi}{6}^- } g(x)

&= limlimits_{x to frac{pi}{6}^- } frac{tanleft(frac{pi}{6}-xright)}{frac{pi}{6}-x} left(frac{pi}{6}-xright)log left(1-2sin xright)

\

&overset{(1)}{=} limlimits_{x to frac{pi}{6}^- } left(frac{pi}{6}-xright)log left(1-2sin xright)

\

&=limlimits_{x to frac{pi}{6}^- } frac{ log (1-2sin x)}{frac{1}{frac{pi}{6}-x}}

\

&overset{mathrm{H}}{=} limlimits_{x to frac{pi}{6}^-} (-2cos x)frac{left(frac{pi}{6}-xright)^2}{1-2sin x}

\

&= -sqrt{3}limlimits_{x to frac{pi}{6}^-} frac{left(frac{pi}{6}-xright)^2}{1-2sin x}

\

&overset{mathrm{H}}{=}

-sqrt{3}limlimits_{x to frac{pi}{6}^-}frac{-2left(frac{pi}{6}-xright)}{-2cos x }

\

&= 0

end{align}$$

where in $(1)$ I have used $lim_{yto0} frac{tan y}{y} = 1$ and $H$ denotes the usage of *L’HΓ΄pital’s rule*.

Hence, we conclude that

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

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 $logleft(frac{sqrt{3}}{2} left(frac{pi}{6} – xright)right) = log frac{sqrt{3}}{2} + log left(frac{pi}{6} – xright)$. 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!

Your job might be simpler if you substitute $pi/6-x=2t$. Then

$$

1-2sin x=1-2sin(pi/6-2t)=1-cos 2t+sqrt{3}sin 2t=2sin t(sin t+sqrt{3}cos t)

$$

Note that in order that the limit makes sense you need $sin x<1/2$, so $0<x<pi/6$ (the lower bound is mostly irrelevant, though), hence $t>0$.

How does this help? You get to evaluate the limit for $tto0$ of

$$

tan2tbigl(log(sin t)+log(2sin t+2sqrt{3}cos t)bigr)

$$

The part $tan2tlog(2sin t+2sqrt{3}cos t)$ poses no problem: its limit is $0$. Then you need to compute the limit of

$$

frac{2cos t}{cos2t}sin tlogsin t

$$

The fraction part has limit $2$. The part $sin tlogsin t$ has limit $0$, as it’s easy to show with l’HΓ΄pital or other methods.

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