# exponential Limit involving Trigonometric function

$$\lim_{x\rightarrow \infty}\left[\cos\left(2\pi\left(\frac{x}{x+1}\right)\right)^{\alpha}\right]^{x^2}$$ where $$\alpha\in \mathbb{Q}$$

Try: $$l=\lim_{x\rightarrow \infty}\bigg[\cos\bigg(2\pi\bigg(1-\frac{1}{x+1}\bigg)\bigg)\bigg]^{x^2}$$

Put $$\displaystyle \frac{1}{x+1}=t.$$ Then limit convert into $$\displaystyle l=\lim_{t\rightarrow 0}\bigg(\cos (2\pi t)^{\alpha}\bigg)^{\frac{1}{t}-1}$$

$$\ln(l)=\lim_{t\rightarrow 1}\bigg(\frac{1}{t}-1\bigg)\ln(\cos(2\pi t)^{\alpha})$$

Could some Help me to solve it, Thanks

Put $$1/x=t$$ to find
$$\lim_{t\to0}\left(\cos\dfrac{2\pi }{t+1}\right)^{1/t^2}$$
$$=\lim_{t\to0}\left(\cos\left(2\pi-\dfrac{2\pi }{t+1}\right)\right)^{1/t^2}$$ as $$\cos(2\pi-x)=\cos x$$
$$\lim_{t\to0}\left(\cos\dfrac{2\pi t}{t+1}\right)^{1/t^2}$$
$$=\left[\lim_{t\to0}\left(1-2\sin^2\dfrac{\pi t}{t+1}\right)^{-\dfrac1{2\sin^2\dfrac{\pi t}{t+1}}}\right]^{-2\lim_{t\to0}\left(\dfrac{2\sin\dfrac{\pi t}{t+1}}t\right)^2}$$