Limit of a sequence: $\left(\cos\left(\frac1n\right)\right)^n$ as $n \to \infty$ How can I solve 
$$\lim_{n\to\infty}\left(\cos\left(\frac1n\right)\right)^n$$
without using calculus?
 A: Assuming $\lim_{u\to \infty} (1-1/u)^u = 1/e,$ we can argue this way: $\cos^2(1/n) + \sin^2(1/n) = 1$ and $\sin (1/n) < 1/n.$ This gives $\cos^2(1/n)>  1-1/n^2.$ Therefore
$$(1-1/n^2)^{n/2} < [\cos (1/n)]^n < 1.$$
The term on the left equals $[(1-1/n^2)^{n^2}]^{1/(2n)}.$ As $n\to \infty,$ the limit inside the brackets is $1/e.$ Raising to the $1/(2n)$ power then gives a limit of $(1/e)^0 = 1.$ Thus the desired limit is $1$ by the squeeze theorem.
A: Something generally like the following?
\begin{align}
\lim_{n \to \infty} \left[\cos\left(\frac{1}{n}\right)\right]^n
    & = \lim_{n \to \infty} \left[1-\frac{1}{2n^2}+o(n^{-3})\right]^n \\
    & = \lim_{n \to \infty} \exp\left[-\frac{1}{2n}+o(n^{-2})\right] \\
    & = 1
\end{align}
A: We must use some standard limits. We have:
$$ \begin{align}
\lim_{n \to \infty} (\cos(1/n))^n &= \lim_{n \to \infty}\exp\left( \dfrac{\cos(1/n) - 1}{1/n} \right) \\
&= \exp(0) = 1
\end{align}$$

 My original answer asserted this: "Using the basic concepts of limits, we imagine inserting a very large value of $n$, say $10^6$. The cosine will spit out a value very close to $1$, but less than one. This value is raised to a very large power, so the value diminishes and tends to zero." This was proved wrong by my own formal calculation and WoframAlpha. I would appreciate some insight into why this loose method didn't work! 
A: $$y = \cos^n (1/n)$$
$$y = \exp\left({\log(\cos (1/n))\over 1/n}\right)$$
$$ \lim_{n \to \infty} y = \lim_{n \to \infty} \exp\left({\log(\cos (1/n))\over 1/n}\right)= \lim_{x \to 0} \exp\left({\log(\cos (x))\over x}\right)-(1)$$
$$ =  \exp\left( {[\log(\cos (x))]'\over [x]'}\large{|}_{x=0}\right)-(2)$$
$$=  \exp\left( {-\tan 0 \over 1}\right) = \exp\left(0\right) = 1$$

$(1)-$ substituing ${1\over n} = x$
$(2)-$ L'Hospital's Rule

