Limit of a sequence

Find the following limit. \[ \lim\limits_{n\rightarrow\infty}⁡\frac 4{2\sqrt{n}-3} \]

Find the limit of the following sequence. \[ {\left({\Bigl (\frac{2n + 1} {n} \Bigr )}^{n}\right)}_{ n=1}^{\infty } \] Hint: The limit of the sequence \({\bigl ({\bigl (1 + \frac{1} {n}\bigr )}^{n}\bigr )}_{n=1}^{\infty }\) is the Euler number \(\mathrm{e}\).

Consider the convergent sequence \[ \left ( \frac{5 - n} {2n - 1}\right )_{n=1}^{\infty } \] and its limit \(L\). Find the index of the first term of the sequence which differs from \(L\) by less than \(\frac{1} {100}\).

Consider the convergent sequence \[ (a_{n})_{n=1}^{\infty } = \left (\frac{4n^{2} + 3n - 250} {2n^{2}} \right )_{n=1}^{\infty } \] and its limit \(L\). Find the maximal difference between \(L\) and the subsequence \((a_{n})_{n=250}^{\infty }\). (In other words, find the maximal difference between \(L\) and the terms of the sequence starting at \(a_{250}\).)