Limit of $\lim\limits_{n\to+\infty}\left(\cos\frac{x}{n} + \sin\frac{2x}{n}\right)^n $

Question

Given the limit: $$ \lim_{n \to \infty}\left[\cos\left(\frac{x}{n}\right) + \sin\left(\frac{2x}{n}\right)\right]^{n} = \alpha $$

Find the value of $\alpha$.

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I suppose there is a special technique of how to solve such expressions. Sadly, I am not familiar with it.

Could anyone put me on point ?.

Answer 1

By using the following Taylor expansions at $0$, $\cos(t)=1+o(t)$, $\sin(t)=t+o(t)$, and $\ln(1+t)=t+o(t)$, we have that $$\begin{align}\left(\cos\left(\frac{x}{n}\right) + \sin\left(\frac{2x}{n}\right)\right)^n &=\exp\left(n \ln\left(\cos\left(\frac{x}{n}\right) + \sin\left(\frac{2x}{n}\right)\right)\right) \\&=\exp\left(n \ln\left(1 + \frac{2x}{n}+o(1/n)\right)\right) \\&=\exp\left(n \left(\frac{2x}{n}+o(1/n)\right)\right). \end{align}$$ Can you take it from here?

Answer 2

Consider the sequence $$a_{n} =\frac{\cos(x/n) +\sin(2x/n)}{1+(2x/n)}$$ and then we have $$n(a_n-1)=n\cdot\frac{\cos(x/n) - 1+\sin(2x/n)-(2x/n)}{1+(2x/n)}$$ and it is easily seen that $n(a_n-1)\to 0$ and therefore $a_n^{n} \to 1$. Using the fact that $(1+(2x/n))^{n}\to e^{2x}$ the desired limit is $e^{2x}$.

Answer 3

Short answer:

$$\cos\frac xn\approx1^*,\\\sin\frac{2x}n\approx\frac{2x}n^*$$ then

$$\left(1+\frac{2x}n\right)^n\to e^{2x}.$$

$^*$ the next terms are negligible.

Answer 4

Since $\sin(2a) =2\cos a\sin a $, $\sin x\sim x$ and $\cos x\sim 1-\frac{x^2}{2}$ $$\lim_{n\to+\infty}\left(\cos\frac{x}{n} + \sin\frac{2x}{n}\right)^n = \lim_{n\to+\infty}\left(1 + 2\sin\frac{x}{n}\right)^n\left(\cos\frac{x}{n} \right)^n \\ \sim \left(1 + \frac{2x}{n}\right)^n\left(1-\frac{x^2}{2n^2}\right)^n \\\overset{\color{red}{h=\frac{x}{n}}}{=}\underbrace{\exp\left(2x\frac{\ln(1+2h)}{2h}\right)}_\overset{h\to 0}{\to e^{2x}} \underbrace{\exp\left(\frac{xh}{2}\frac{\ln(1-\frac{h^2}{2})}{\frac{h^2}{2}}\right)}_{\overset{h\to 0}{\to 1} }\overset{h\to 0}{\to} e^{2x}$$

Given that $$ \lim_{X\to 0}\frac{\ln(1+X)}{X} =0$$