By the Squeeze Theorem, since , you have .
To show that , you can use an proof. Given , can you show that there exists such that for all , and ? First, let's use an induction argument to show that for all .
Base of induction: . Then is true. . Then is also true.
Assume the proposition is true for some positive integer (we already showed it is true for , so this assumption is true). Then . For all , . So, (the second inequality came from application of the induction assumption). Hence, for all .
Now that is proven, choose any positive integer . Then for any , we have , which implies . Dividing both sides by 2 gives . Taking the reciprocal of both sides flips the inequality: . This is exactly what you wanted to prove.
There are various other ways of proving it.