if $\lim x_n = x$ then show $\lim \inf(x_n) = x$ if $\lim x_n = x$ then show $\lim \inf(x_n) = x$
an attempt:
we have $x_n \rightarrow x$ so $x_n$ is bounded, hence there exists an infimum $\inf x_n$, by defn of $\inf x_n$ $\inf x_n \leq x_n$  so $\inf x_n - x \leq x_n - x < \epsilon$ i.e. $\inf x_n - x < \epsilon$ but I'm not sure how to show it is also > $-\epsilon$ 
edit: could we possibly say that since $\inf(x_n) + \epsilon$ is not a lower bound, there exits a $N$ such that $ x_N < inf(x_N) + \epsilon$, hence $-\epsilon < x_N - x< \inf(x_N) + \epsilon - x $ i.e. $-\epsilon < 0 < \inf(x_N) - x$ then choose $n>N$ and we're done?
 A: If sequence $\left(x_{n}\right)$ is bounded and $y_{n}:=\inf_{k\geq n}x_{k}$
then we recognize in $\left(y_{n}\right)$ a non-decreasing and bounded sequence
with $y_{n}\leq x_{n}$. Such a sequence has automatically a limit $y$ and in fact $y:=\liminf x_n$.
If moreover $\left(x_{n}\right)$ converges with $\lim_{n\rightarrow\infty}x_{n}=x$
then $y\leq x$ as a consequence of $y_{n}\leq x_{n}$ for each $n$. 
Also for every $\varepsilon>0$ some $n\left(\varepsilon\right)$
exists with $n\geq n_{\varepsilon}\Rightarrow x_{n}>x-\varepsilon$
so that $y\geq y_{n\left(\varepsilon\right)}\geq x-\varepsilon$.
This allows the conclusion $y\geq x$. Proved is now that: $$\lim_{n\rightarrow\infty}\inf_{k\geq n}x_{k}=\lim_{n\rightarrow\infty}x_{n}$$
$\liminf x_{n}$ is a short notation for $\lim_{n\rightarrow\infty}\inf_{k\geq n}x_{k}$
A: If we have a sequence $x_n = (1 + \frac 1n)^n, n \in \mathbb N$, then $\lim_{n \to \infty}x_n = e$, but $\inf_{n \in \mathbb N}(x_n) = 2, \forall n \in \mathbb N$. Isn't it a counterexample?
