Is there a way to compute $\lim\limits_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}$ without using L'hopital?

Question

I can compute $$\lim_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}$$ using L'hopital and the limit equals $\frac{\sqrt{3}}{12}$, but is there another way to compute this limit without using L'hopital, please do it for me if there is.

Many thanks.

Answer 1

$$\lim_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}=\lim_{x\to\pi/3}\frac{2\cos x-1}{\ln(1+\sin3x)}\frac{1}{\sqrt{3+2\cos x}+2}$$ $$=\frac{1}{4}\lim_{x\to\pi/3}\frac{2\cos x-1}{\ln(1+\sin3x)}=\frac{1}{4}\lim_{x\to\pi/3}\frac{2\cos x-1}{\sin(3x)}\frac{\sin(3x)}{\ln(1+\sin3x)}$$

$\lim_{x\to\pi/3}\frac{2\cos x-1}{\sin(3x)}$ is just a standard trig limit, while

$$\lim_{x\to\pi/3}\frac{\sin(3x)}{\ln(1+\sin3x)}=\frac{1}{\lim_{x\to\pi/3}\frac{\ln(1+\sin3x)}{\sin(3x)}}=\frac{1}{\lim_{y \to 1}\frac{\ln(y)}{y-1}}=\frac{1}{\ln'(1)}=1$$

Answer 2

Let's replace $x$ by $x+\pi/3$ and let $x \to 0$. I will be as simple-minded as I can.

We use $\sin(\pi/3) = \sqrt{3}/2$ and $\cos(\pi/3) = 1/2$.

$\cos(x+\pi/3) = \cos(x) \cos(\pi/3) - \sin(x)\sin(\pi/3) = \cos(x)/2 - \sin(x)\sqrt{3}/2 $. As $x \to 0$, $\cos(x+\pi/3) \to 1/2$.

$\sin(3(x+\pi/3)) = \sin(3x+3\pi/3) = \sin(3x+\pi) = \sin(3x)\cos(\pi) + \cos(3x)\sin(\pi) = -\sin(3x) \to 0 $.

So, initially, this goes to $0/0$.

Let's use $\sin(x) \approx x$ and $\cos(x) \approx 1-x^2/2$ as $x \to 0$.

$\cos(x+\pi/3) \approx (1-x^2/2)/2 - x\sqrt{3}/2 \approx 1/2- x\sqrt{3}/2 $ and $\sin(3(x+\pi/3)) \approx -3x$.

Then $\sqrt{3+2\cos (x+\pi/3}-2 \approx \sqrt{3+2(1/2- x\sqrt{3}/2)} =\sqrt{4-x\sqrt{3}} =2\sqrt{1-x\sqrt{3}/4} \approx 2(1-x\sqrt{3}/8) =2-x\sqrt{3}/4 $ so the numerator is about $-x\sqrt{3}/4$ as $x \to 0$.

For the denominator, we have $\ln(1+x) \approx x$ as $x \to 0$, so $\ln(1+\sin(3(x+\pi/3)) \approx \ln(1-3x) \approx -3x$

so the ratio is $\frac{-x\sqrt{3}/4}{-3x} =\frac{\sqrt{3}}{12} $.

This can be made rigorous if we can use $\sin(x) = x +o(x^2)$ or $\sin(x) \approx x +O(x^3)$ as $x \to 0$.

Answer 3

We will have to use some limiting fact about $\log$. We will use that $\displaystyle\lim_{u\to0}\frac{\log(1+u)}{u}=1$. $$ \begin{align} \lim_{x\to\pi/3}\frac{\sqrt{3+2\cos(x)}-2}{\log(1+\sin(3x))} &=\lim_{x\to\pi/3}\frac{\color{#00A000}{\sqrt{3+2\cos(x)}-2}}{\log(1+\sin(3x))} \frac{\color{#00A000}{\sqrt{3+2\cos(x)}+2}}{\sqrt{3+2\cos(x)}+2}\\ &=\lim_{x\to\pi/3}\frac{\color{#00A000}{2\cos(x)-1}}{\log(1+\sin(3x))} \color{#C00000}{\frac1{\sqrt{3+2\cos(x)}+2}}\\ &=\color{#C00000}{\frac14}\lim_{x\to\pi/3}\frac{\color{#0000FF}{2\cos(x)-1}}{\log(1+\sin(3x))} \frac{\color{#0000FF}{(2\cos(x)+1)\sin(x)}}{(2\cos(x)+1)\sin(x)}\\ &=\frac14\lim_{x\to\pi/3}\frac{\color{#0000FF}{\sin(3x)}}{\log(1+\sin(3x))} \frac1{\color{#00A000}{(2\cos(x)+1)\sin(x)}}\\ &=\frac1{4\color{#00A000}{\sqrt3}}\lim_{x\to\pi/3}\frac{\sin(3x)}{\log(1+\sin(3x))}\\ &=\frac1{4\sqrt3}\lim_{u\to0}\frac{u}{\log(1+u)}\\ &=\frac1{4\sqrt3} \end{align} $$

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Proof of limit used above

Using $\log(x)=\lim\limits_{n\to\infty}n(x^{1/n}-1)$, which simply inverts $e^x=\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n$, we have $$ \begin{align} \frac{\log(1+u)}{u} &=\lim_{n\to\infty}n\frac{(1+u)^{1/n}-1}{(1+u)-1}\\ &=\lim_{n\to\infty}\frac{n}{(1+u)^{(n-1)/n}+(1+u)^{(n-2)/n}+\dots+1} \end{align} $$ which is easily between $\frac1{1+u}$ and $1$. Thus, the Squeeze Theorem yields $$ \lim_{u\to0}\frac{\log(1+u)}{u}=1 $$