# Finding the limit of a sequence using squeeze theorem

How would one use the squeeze theorem to find the limit $$\lim_{n\rightarrow\infty} (\frac{1+n}{2n})^n$$ ?

Unlike functions that incorporate sine or cosine, it's less apparent to me what the bounds should be.

Take the bounds: $\frac{1}{2}<\frac{1+n}{2n}<\frac{\frac{n}{2}+n}{2n}=\frac{3}{4}$. They show that the limit should be $0$.
The sequence consists of positive values, so we have $$0\le (\frac{1+n}{2n})^n=(\frac{1}{2})^n\cdot (1+\frac{1}{n})^n\le (\frac{1}{2})^n\cdot e$$
This immediately shows that the limit is $0$.
When $n>1$, $\frac{1+n}{2n} \le \frac{3}{4}$