find limit of a sequence I have a sequence:
$$ X_n\:=\:\cos\left(\left(\frac{3^n+\pi^n}{3^n+\left(\pi-\frac{1}{4} \right)^n} \right)^{1/n}\right) $$
I have to find the limit when $n \to \infty $ where $n \in \mathbb{N}$.
Which is the best way to find the answer ? Can I reduce or use the Squeeze theorem in that case ?
 A: $${3^n+\pi^n\over3^n+(\pi-{1\over4})^n}=\left(\pi\over3\right)^n{1+\left(3\over\pi\right)^n\over1
+\left(\pi-{1\over4}\over3\right)^n}$$
Now $\pi-{1\over4}\lt3\lt\pi$, so $1+(3/\pi)^n\to1$ and $1+((\pi-{1\over4})/3)^n\to1$.  Thus
$$\left(3^n+\pi^n\over3^n+(\pi-{1\over4})^n\right)^{1/n}\to{\pi\over3}$$
Taking the cosine gives the limit $\cos(\pi/3)=1/2$.
A: HINT: write your term as $$\cos\left(\frac{1+\left(\frac{\pi}{3}\right)^n}{1+\left(\frac{\pi-\frac{1}{4}}{3}\right)^n}\right)^{1/n}$$
A: The answer is $\frac12$, because$$\sqrt[n]{\frac{3^n+\pi^n}{3^n+\left(\pi-\frac14\right)^n}}$$behaves as$$\sqrt[n]{\frac{\pi^n}{3^n}},$$that is, as $\frac\pi3$, when $n$ is very large. And $\cos\left(\frac\pi3\right)=\frac12$.
A: Almost as a reflex, I'd go for Taylor expansions. (Because it works. Not always the most elegant, bu at least it gets us somewhere.) When $n\to\infty$, letting $x\stackrel{\rm def}{=} \frac{\pi -\frac{1}{4}}{3}$ and $y\stackrel{\rm def}{=} \frac{3}{\pi}$, with $x>y$ (so that $y^n=o(x^n)$)
\begin{align}
\cos\left(\left(\frac{3^n+\pi ^n}{3^n+\left(\pi -\frac{1}{4}\right)^n}\right)^{\frac{1}{n}}\right)
&= \cos\left(\frac{\pi}{3}\left(\frac{1+\left(\frac{3}{\pi}\right)^n}{1+\left(\frac{\pi -\frac{1}{4}}{3}\right)^n}\right)^{\frac{1}{n}}\right)\\
&= \cos\left(\frac{\pi}{3}\left(\left(1+y^n\right)\left(1+x^n + o(x^{n})\right)\right)^{\frac{1}{n}}\right)\\
&= \cos\left(\frac{\pi}{3}\left(1+y^n + x^n +o(x^n)\right)^{\frac{1}{n}}\right) \\
&= \cos\left(\frac{\pi}{3}\left(1+x^n +o(x^n)\right)^{\frac{1}{n}}\right) \tag{as $x>y$}\\
&= \cos\left(\frac{\pi}{3}\exp\left(\frac{1}{n}\ln\left(1+x^n +o(x^n)\right)\right)\right) \\
&= \cos\left(\frac{\pi}{3}\exp\left(\frac{x^n}{n}+o\left(\frac{x^n}{n}\right)\right)\right) \\
&= \cos\left(\frac{\pi}{3}\left(1+\frac{x^n}{n}+o\left(\frac{x^n}{n}\right)\right)\right) \\
&\xrightarrow[n\to\infty]{} \cos\frac{\pi}{3} = \frac{1}{2}
\end{align}
A: Denote the argument of the cosine as $a_n$. Estimate:
$$\left( \frac{\pi ^n \left(\left(\frac{3}{\pi}\right)^n+1\right)}{2\cdot 3^n}\right)^{1/n}<a_n<\left(\frac{2\cdot \pi ^n}{3^n \left(1+\left(\frac{\pi -\frac14}{3}\right)^n\right)}\right)^{1/n}.$$
Taking limit:
$$\frac{\pi }{3}\le \lim_{n\to\infty} a_n \le \frac{\pi}{3}.$$
Hence:
$$\lim_{n\to\infty} \cos{a_n} =\cos{\frac{\pi}{3}}=\frac12.$$
A: $$a_n=\sqrt[n]{\frac{3^n+ \pi^n}{3^n+ (\pi -1/4)^n}}\leq \sqrt[n]{\frac{2 \pi^n}{3^n}}=\sqrt[n]{2} \frac{\pi}{3} \to \frac{\pi}{3}$$
Also $$a_n=\sqrt[n]{\frac{3^n+ \pi^n}{3^n+ (\pi -1/4)^n}} \geq \sqrt[n]{ \frac{\pi^n}{3^n+3^n}}=\frac{1}{\sqrt[n]{2}} \frac{\pi}{3} \to \frac{\pi}{3}$$
So from squeeze theorem $a_n \to \frac{\pi}{3}$.
Now using the continuity of $\cos{x}$ we have that $$\cos{a_n} \to \cos{\frac{\pi}{3}}=\frac{1}{2}$$
