Csdrhrt

Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.

But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.

Put $t= x- dfrac{π}{3}$

$lim limits_{x to frac{π}{3}} dfrac{2 sin x - sqrt{3}}{cos frac{3x}{2}}$

$= limlimits_{t to 0} dfrac{2 sin left(t+frac{π}{3} right) - sqrt{3}}{cos left( frac{3t}{2} + frac{π}{2}right)}$

$= limlimits_{t to 0} dfrac{sqrt{3}- sin t - sqrt{3} cos t}{ sin frac{3t}{2}}$

Where do I go from here, I'm not able to eliminate the $t$ fully from the Nr and Dr, any help?

Or any other alternative way that uses only the fact that $limlimits_{x to 0} dfrac{ sin x}{x} = 1$?

Thanks :)

So we already have, after putting $;t:=x-fracpi3;$ :

$$dfrac{2 sin left(t+frac{π}{3} right) - sqrt{3}}{cos left( frac{3t}{2} + frac{π}{2}right)}=frac{2left(sin tcdotfrac12+frac{sqrt3}2cos tright)-sqrt3}{-sinfrac{3t}2}=-frac{sin t}{sinfrac{3t}2}-sqrt3frac{cos t -1}{sinfrac{3t}2}=$$$${}$$

$$=-frac23cdotfrac{frac{3t}2}{sinfrac{3t}2}cdotfrac{sin t}t+sqrt3cdotfrac23cdotfrac{frac{3t}2}{sinfrac{3t}2}cdotfrac{1-cos t}txrightarrow[tto0]{}-frac23cdot1cdot1+frac2{sqrt3}cdot1cdot0=-frac23$$

$lim limits_{x to frac{π}{3}} dfrac{2 sin x - sqrt{3}}{cos frac{3x}{2}}=2lim limits_{x to frac{π}{3}} dfrac{sin x - sin cfrac pi 3}{cos frac{3x}{2}}$

$=4lim limits_{x to frac{π}{3}} dfrac{cos (cfrac x2+cfrac pi 6) sin (cfrac x2 - cfrac pi 6)}{sin (frac pi 2-frac {3x}{2})}$

$=4lim limits_{x to frac{π}{3}} dfrac{cos (cfrac x2+cfrac pi 6) (cfrac x2 - cfrac pi 6)}{frac pi 2-frac {3x}{2}}$

$=-4/3lim limits_{x to frac{π}{3}} {cos (cfrac x2+cfrac pi 6) }$

$=-4/3cos pi/3=-2/3$

To end your computation, you can split your expression in two:
$$frac{sqrt{3}- sin t - sqrt{3} cos t}{ sin frac{3t}{2}} =-frac{sin t}{ sin frac{3t}{2}}+sqrt 3,frac{1- cos t}{sin frac{3t}{2}} $$
Now it is standard that $;lim_{tto 0}dfrac{sin at}{sin bt}=dfrac ab$. On the other hand
$$frac{1-cos t}{sinfrac{3t}{2}}=frac{2sin^2t}{sinfrac{3t}{2}}=2sin t,frac{sin t}{sinfrac{3t}{2}}.$$
Can you conclude now?

Just to give another approach, let's let $f(x)=2sin x$ and $g(x)=cos(3x/2)$ and use the fact(s) that $f'(x)=2cos x$, $g'(x)=-{3over2}sin(3x/2)$, $f(pi/3)=2sin(pi/3)=sqrt3$, and $g(pi/3)=cos(pi/2)=0$. From the definition of the derivative, it follows that

$$lim_{xtopi/3}{2sin x-sqrt3over x-pi/3}=lim_{pi/3}{f(x)-f(pi/3)over x-pi/3}=f'(pi/3)=2cos(pi/3)=1$$

and

$$lim_{xtopi/3}{cos(3x/2)over x-pi/3}=lim_{xtopi/3}{g(x)-g(pi/3)over x-pi/3 }=g'(pi/3)=-{3over2}sin(pi/2)=-{3over2}$$

Thus

$$lim_{xtopi/3}{2sin x-sqrt2overcos(3x/2)}={displaystylelim_{xtopi/3}{2sin x-sqrt3over x-pi/3}overdisplaystylelim_{xtopi/3}{cos(3x/2)over x-pi/3}}={1over-{3over2}}=-{2over3}$$

$$limlimits_{t to 0} dfrac{sqrt{3}- sin t - sqrt{3} cos t}{ sin frac{3t}{2}}=limlimits_{t to 0} dfrac{sqrt{3}- sin t - sqrt{3} cos t}{ sin frac{3t}{2}}cdotfrac{3t/2}{3t/2}$$
$$=frac{2}{3} limlimits_{t to 0} dfrac{sqrt{3}- sin t - sqrt{3} cos t}{ t} (1)$$
Now, $1-cost=2sin^2(t/2)$
$$limlimits_{t to 0} dfrac{sqrt{3}- sin t - sqrt{3} cos t}{t}= limlimits_{t to 0}( frac{sqrt{3}(1-cos t)}{t}-frac{sint}{t})$$
$$=limlimits_{t to 0}( frac{2sin^2(t/2)}{t}-frac{sint}{t})$$
$$=limlimits_{t to 0}( frac{2sin^2(t/2)}{t^2/4}frac{t^2/4}{t}-frac{sint}{t})$$
$$=0-1=-1$$
Put -1 in (1) to get the final limit, $-2/3$.