Tìm các giới hạn sau:

a) $\mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2} + n + 1}}{{2{n^2} + 1}}$;

b) $\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} + 2n} - n} \right)$.

Lời giải:

a) $\mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2} + n + 1}}{{2{n^2} + 1}}$$= \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left( {1 + \frac{1}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {2 + \frac{1}{{{n^2}}}} \right)}}$$= \mathop {\lim }\limits_{n \to + \infty } \frac{{1 + \frac{1}{n} + \frac{1}{{{n^2}}}}}{{2 + \frac{1}{{{n^2}}}}} = \frac{1}{2}$.

b) $\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} + 2n} - n} \right)$$= \mathop {\lim }\limits_{n \to + \infty } \frac{{\left( {{n^2} + 2n} \right) - {n^2}}}{{\sqrt {{n^2} + 2n} + n}}$

$= \mathop {\lim }\limits_{n \to + \infty } \frac{{2n}}{{\sqrt {{n^2}\left( {1 + \frac{2}{n}} \right)} + n}}$$= \mathop {\lim }\limits_{n \to + \infty } \frac{{2n}}{{n\sqrt {1 + \frac{2}{n}} + n}}$

$= \mathop {\lim }\limits_{n \to + \infty } \frac{{2n}}{{n\left( {\sqrt {1 + \frac{2}{n}} + 1} \right)}}$$= \mathop {\lim }\limits_{n \to + \infty } \frac{2}{{\sqrt {1 + \frac{2}{n}} + 1}} = \frac{2}{{\sqrt 1 + 1}} = 1$.