Calculating $\lim_{x\to\pi} \frac{1}{x-\pi}\left(\sqrt{\frac{4\cos²x}{2+\cos x}}-2\right)$

Question

I'm trying to calculate the following limit:

$$\lim_{x\to\pi} \dfrac{1}{x-\pi}\left(\sqrt{\dfrac{4\cos²x}{2+\cos x}}-2\right)$$

I thought of calculating this:

$$\lim_{t\to0} \dfrac{1}{t}\left(\sqrt{\dfrac{4\cos²(t+\pi)}{2+\cos(t+\pi)}}-2\right)$$

Which is the same as:

$$\lim_{t\to0} \dfrac{1}{t}\left(\sqrt{\dfrac{4\cos²t}{2-\cos t}}-2\right)$$

I don't have an idea about where to go from here.

Answer 1

Note that $a-b = \frac{a^2 -b^2}{a+b}$. Then:

$$ \frac{1}{t}\left(\sqrt{\frac{4\cos²t}{2-\cos t}}-2\right) = \frac{1}{t}\left(\frac{\frac{4\cos²t}{2-\cos t}-4}{\sqrt{\frac{4\cos²t}{2-\cos t}}+2}\right) = \frac{1}{t}\left(\frac{\frac{4\cos²t-8 + 4 \cos t}{2-\cos t}}{\sqrt{\frac{4\cos²t}{2-\cos t}}+2}\right) = \frac{\frac{4\cos²t-8 + 4 \cos t}{t(2-\cos t)}}{\sqrt{\frac{4\cos²t}{2-\cos t}}+2} $$

The limit of the denominator is easy, so we just need to calculate

$$ \lim_{t \to 0} \frac{4\cos²t-8 + 4 \cos t}{t(2-\cos t)} = \lim_{t \to 0} \frac{4 (\cos t + 2)(\cos t - 1)}{t(2-\cos t)} = 4\cdot 3 \cdot \lim_{t \to 0}\frac{\cos t - 1}{t} = 0 $$

Answer 2

Let $f(x)=\sqrt{\dfrac{4\cos^{2}x}{2+\cos x}}$, the limit $L$ is actually the derivative of $f$ at $x=\pi$, so $$L=\dfrac{1}{2}\left(\dfrac{4\cos^{2}x}{2+\cos x}\right)^{-1/2}\dfrac{(2+\cos x)(-8\sin x)-4(\cos^{2}x)(-\sin x)}{(2+\cos x)^{2}}\bigg|_{x=\pi}=\cdots$$.

Answer 3

From here by first order binomial expansion

$$\frac{1}{t}\left(\sqrt{\frac{4\cos²t}{2-\cos t}}-2\right)=\frac1t(2\cos t(1-(1-\cos t))^{-\frac12}-2)\sim\frac1t(2\cos t(1+\frac12(1-\cos t))-2)=\frac1t(2\cos t+\cos t-\cos^2t-2)=\frac{-\cos^2t+3\cos t-2}{t}=\frac{(\cos t-1)(2-\cos t)}{t^2}\cdot t\to -\frac12 \cdot 0=0$$

As an alternative by algebraic manipulation

$$\frac{1}{t}\left(\sqrt{\frac{4\cos^2t}{2-\cos t}}-2\right)= \frac{1}{t}\frac{\sqrt{4\cos^2t}-2\sqrt{2-\cos t}}{\sqrt{2-\cos t}} \frac{\sqrt{4\cos^2t}+2\sqrt{2-\cos t}}{\sqrt{4\cos^2t}+2\sqrt{2-\cos t}} =\frac{1}{t}\frac{4\cos^2t-8+4\cos t}{\sqrt{2-\cos t}(\sqrt{4\cos^2t}+2\sqrt{2-\cos t})} =t\frac{\cos t -1}{t^2}\frac{4(\cos t+2)}{\sqrt{2-\cos t}(\sqrt{4\cos^2t}+2\sqrt{2-\cos t})}\to0\cdot-\frac12\cdot 2=0$$

Answer 4

$$ \begin{aligned} \lim _{x\to \pi }\:\frac{1}{x-\pi }\left(\sqrt{\frac{4\cos^²x}{2+\cos x}}-2\right) &= \lim _{y\to 0}\:\frac{1}{\left(y+\pi \right)-\pi }\left(\sqrt{\frac{4\cos^²(y+\pi)}{2+\cos(y+\pi)}}-2\right) \\ &= \lim _{y\to 0} \frac{2\cos y-2\sqrt{-\cos y+2}}{y\sqrt{-\cos y+2}} \\ &= \lim _{y\to 0} \frac{2\left(1-\frac{y^2}{2!}+o\left(y^2\right)\right)-2\left(1+\frac{y^2}{4}+o\left(y^2\right)\right)}{y\left(1+\frac{y^2}{4}+o\left(y^2\right)\right)} \\ &= \lim _{y\to \:0}-\frac{6y+o(y^2)}{4+y^2+o(y^3)} \\ &= \color{red}{0} \end{aligned} $$ Solved with Taylor expansion