Tính lim(2n + căn bậc hai (4n^2 - 2n + 1))
Tính \(\lim \left( {2n + \sqrt {4{n^2} - 2n + 1} } \right)\).
\(\lim \left( {2n + \sqrt {4{n^2} - 2n + 1} } \right)\)
= \(\lim \left[ {2n + \sqrt {{n^2}\left( {4 - \frac{2}{n} + \frac{1}{{{n^2}}}} \right)} } \right]\)
= \(\lim \left[ {2n + n\sqrt {\left( {4 - \frac{2}{n} + \frac{1}{{{n^2}}}} \right)} } \right]\)
= \(\lim n\left( {2 + \sqrt {4 - \frac{2}{n} + \frac{1}{{{n^2}}}} } \right)\)
Ta có: lim n = +∞
lim \(\lim \left( {2 + \sqrt {4 - \frac{2}{n} + \frac{1}{{{n^2}}}} } \right) = 2 + \sqrt {4 - 0 + 0} = 4\)
Suy ra: \(\lim n\left( {2 + \sqrt {4 - \frac{2}{n} + \frac{1}{{{n^2}}}} } \right)\) = +∞ . 4 = +∞.