Question is:
$$\lim_{x\to \infty} (x+1) \cos \left(2\cos ^{-1}(\frac{x}{x + 1}) (a - \frac{1}{2})\right) - x$$
The answer is $4x-4x^2$ but I'm not sure how to get there. I have to use expansions of the cosine function but I have tried the Taylor (and MacLaurin) series and they end up canceling out unless I've done something wrong.
Note: I've seen a similar question on here before but the answer to the question does not mention any expansions and this is the part I am having trouble with.
Let $t = \frac{x}{x+1}$
$$(x+1) \cos \left(2\cos ^{-1}(\frac{x}{x + 1}) (a - \frac{1}{2})\right) - x = \frac{\cos \left(2\cos ^{-1}(t) (a - \frac{1}{2})\right)}{1-t} - \frac{t}{1-t} = \frac{\cos \left(2\cos ^{-1}(t) (a - \frac{1}{2})\right) -t}{1-t} $$ If $x\to \infty$ then $t \to 1^{-}$. So we have
$$L = \lim_{x \to \infty}(x+1) \cos \left(2\cos ^{-1}(\frac{x}{x + 1}) (a - \frac{1}{2})\right) - x =\lim_{t \to 1^{-}}\frac{\cos \left(2\cos ^{-1}(t) (a - \frac{1}{2})\right) -t}{1-t} $$ Also $$\lim_{t \to 1^{-}}{\cos \left(2\cos ^{-1}(t) (a - \frac{1}{2})\right) -t} = \lim_{t \to 1^{-}} 1 -t = 0$$ So we are allowed to use L'Hôpital's rule. Let $\alpha =2(a-\frac{1}{2})$ $$\lim_{t \to 1^{-}}\frac{\cos(\alpha\cos ^{-1}(t)) -t}{1-t} = \lim_{t \to 1^{-}} 1 -\frac{\alpha\sin(\alpha\cos ^{-1}(t))}{\sqrt{1-t^2}}$$ Applying again L'Hôpital's rule $$\lim_{t \to 1^{-}} \frac{\sin(\alpha\cos ^{-1}(t))}{\sqrt{1-t^2}} = \lim_{t \to 1^{-}}\frac{\cos(\alpha\cos ^{-1}(t))}{t(1-t^2)^\frac{-1}{2}}\frac{\alpha}{\sqrt{1-t^2}} = \alpha$$ So the answer is $$L = 1- \alpha^2 = 1 - 4(a-\frac{1}{2})^2 = -4a(a-1)$$ This is confirmed by WA.
It's also possible to compute the limit using only trigonometry. Let $y =\alpha\cos ^{-1}(t)$ $$\lim_{t \to 1^{-}}\frac{\cos(\alpha\cos ^{-1}(t)) -t}{1-t} = \lim_{y \to 0^{+}} \frac{\cos(y) - \cos(\frac{y}{\alpha})}{1 - \cos(\frac{y}{\alpha})} = \lim_{y \to 0^{+}}\frac{-2\sin(\frac{y}{2} + \frac{y}{2\alpha})\sin(\frac{y}{2} - \frac{y}{2\alpha})}{2\sin^2(\frac{y}{2\alpha})} = \ \ -\lim_{y \to 0^{+}}\frac{\sin(\frac{y}{2} + \frac{y}{2\alpha})}{\sin(\frac{y}{2\alpha})} \times \lim_{y \to 0^{+}}\frac{\sin(\frac{y}{2} - \frac{y}{2\alpha})}{\sin(\frac{y}{2\alpha})}$$
Let $z = \frac{y}{2\alpha}$, multiply and divide by $z$ then separate two limits
$$\lim_{y \to 0^{+}}\frac{\sin(\frac{y}{2} + \frac{y}{2\alpha})}{\sin(\frac{y}{2\alpha})} = \lim_{z \to 0^{+}} \frac{\sin(\alpha z + z)}{\sin(z)} = \lim_{z \to 0^{+}}\frac{z}{\sin(z)} \times \lim_{z \to 0^{+}}\frac{\sin(\alpha z + z)}{z} = \alpha + 1$$ The first limit is $1$ and for the second one let $p = \alpha z + z$ $$\lim_{z \to 0}\frac{\sin(\alpha z + z)}{z} = \lim_{p \to 0} (1+\alpha)\frac{\sin(p)}{p} = \alpha+1$$ Similarly $$\lim_{y \to 0^{+}}\frac{\sin(\frac{y}{2} - \frac{y}{2\alpha})}{\sin(\frac{y}{2\alpha})} = \alpha - 1$$ And again $L = -(\alpha+1)(\alpha - 1) = 1 - \alpha^2$