Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\lim \left( {\sqrt {{n^2} + 1} - n} \right)\\
= \lim \frac{{\left( {\sqrt {{n^2} + 1} - n} \right)\left( {\sqrt {{n^2} + 1} + n} \right)}}{{\sqrt {{n^2} + 1} + n}}\\
= \lim \frac{{\left( {{n^2} + 1} \right) - {n^2}}}{{\sqrt {{n^2} + 1} + n}}\\
= \lim \frac{1}{{\sqrt {{n^2} + 1} + n}}\\
\lim \sqrt {{n^2} + 1} + n = + \infty \Rightarrow \lim \frac{1}{{\sqrt {{n^2} + 1} + n}} = 0\\
b,\\
\lim \left( {\sqrt {{n^2} + 4n} - n} \right)\\
= \lim \frac{{\left( {\sqrt {{n^2} + 4n} - n} \right)\left( {\sqrt {{n^2} + 4n} + n} \right)}}{{\sqrt {{n^2} + 4n} + n}}\\
= \lim \frac{{\left( {{n^2} + 4n} \right) - {n^2}}}{{\sqrt {{n^2} + 4n} + n}}\\
= \lim \frac{{4n}}{{\sqrt {{n^2} + 4n} + n}}\\
= \lim \frac{4}{{\sqrt {1 + \frac{4}{n}} + 1}}\\
= \frac{4}{{\sqrt 1 + 1}}\\
= 2
\end{array}\)
