Đáp án:
\(\begin{array}{l}
5)\quad \lim\dfrac{\left(2n\sqrt n + 1\right)\left(3 + \sqrt n\right)}{(n+1)^2}= 2\\
6)\quad \lim\left(1 + n^2 - \sqrt{n^4 + 3n+1}\right)= 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
5)\quad \lim\dfrac{\left(2n\sqrt n + 1\right)\left(3 + \sqrt n\right)}{(n+1)^2}\\
= \lim\dfrac{\left(2 + \dfrac{1}{n\sqrt n}\right)\left(\dfrac{3}{\sqrt n} + 1\right)}{\left(1 + \dfrac1n\right)^2}\\
= \dfrac{(2 + 0)(0+1)}{(1+0)^2}\\
= 2\\
6)\quad \lim\left(1 + n^2 - \sqrt{n^4 + 3n+1}\right)\\
= \lim\dfrac{\left(1 + n^2 - \sqrt{n^4 + 3n+1}\right)\left(1 + n^2 + \sqrt{n^4 + 3n+1}\right)}{1 + n^2 + \sqrt{n^4 + 3n+1}}\\
= \lim\dfrac{2n^2 - 3n}{1 + n^2 + \sqrt{n^4 + 3n+1}}\\
= \lim\dfrac{2 - \dfrac3n}{\dfrac{1}{n^2} + 1 + \sqrt{1 + \dfrac{3}{n^3} + \dfrac{1}{n^4}}}\\
= \dfrac{2 - 0}{0 + 1 + \sqrt{1 + 0 + 0}}\\
= 1
\end{array}\)
