Đáp án:
$\begin{array}{l}a)\quad \lim\dfrac{2n^2 - n + 3}{n^2 + 3n+1} = 2\\b) \quad \lim\dfrac{n^2 - 4n + 1}{2n^2 - 3n + 1} = \dfrac12\\c)\quad \lim\dfrac{4n^4 - n^2 + 3}{(2n+1)(3-n)(n^2 + 2)} = -2\\d)\quad \lim\dfrac{(n^2 + 2)(n-2)^2}{(n+1)(2n+3)^3} = \dfrac18\\ e)\quad \lim\dfrac{2n^2 - n + 3}{n^3 + 4n^2 + 1} = 0\\ f)\quad \lim\dfrac{2n\sqrt n - 4n + 2}{n^2 + 3n + 1} =0\\g) \quad \lim\dfrac{n\sqrt n - n + 2}{n^2 + n\sqrt{3n} + 1} =0\\ h)\quad \lim\dfrac{n^4 - 5n^2 + 4}{n^2 - 3n + 1}=+\infty\\ i) \quad \lim\dfrac{2n^3 - 3n + 3}{-n^2 + 3n - 2} = -\infty\\ j) \quad \lim\dfrac{n(n^2 + 2n)}{3 - 2n + 2n^2} = +\infty \end{array}$
Giải thích các bước giải:
$\begin{array}{l}a)\quad \lim\dfrac{2n^2 - n + 3}{n^2 + 3n+1}\\ = \lim\dfrac{2 - \dfrac1n + \dfrac{3}{n^2}}{1 + \dfrac3n +\dfrac{1}{n^2}}\\ = \dfrac{2 -0 + 0}{1 + 0 + 1}\\ = 2\\ b)\quad \lim\dfrac{n^2 - 4n + 1}{2n^2 - 3n + 1}\\ = \lim\dfrac{1 - \dfrac4n + \dfrac{1}{n^2}}{2 - \dfrac3n + \dfrac{1}{n^2}}\\ = \dfrac{1 - 0 + 0}{2 - 0 + 0}\\ = \dfrac12\\ c)\quad \lim\dfrac{4n^4 - n^2 + 3}{(2n+1)(3-n)(n^2 + 2)}\\ = \lim\dfrac{4 - \dfrac{1}{n^2} + \dfrac{3}{n^4}}{\left(2 +\dfrac1n\right)\left(\dfrac3n - 1\right)\left(1 + \dfrac{2}{n^2}\right)}\\ = \dfrac{4 - 0 + 0}{(2 +0)(0 - 1)(1 + 0)}\\ =-2\\ d)\quad \lim\dfrac{(n^2 + 2)(n-2)^2}{(n+1)(2n+3)^3}\\ = \lim\dfrac{\left(1 + \dfrac{2}{n^2}\right)\left(1 - \dfrac2n\right)^2}{\left(1 + \dfrac1n\right)\left(2 + \dfrac3n\right)^3}\\ = \dfrac{(1 +0)(1 -0)^2}{(1 + 0)(2 + 0)^3}\\ =\dfrac18\\ e)\quad \lim\dfrac{2n^2 - n + 3}{n^3 + 4n^2 + 1}\\ = \lim\dfrac{\dfrac2n - \dfrac{1}{n^2} + \dfrac{3}{n^3}}{1 + \dfrac4n + \dfrac{1}{n^3}}\\ = \dfrac{0 - 0 + 0}{1 + 0 + 0}\\ = 0\\ f)\quad \lim\dfrac{2n\sqrt n - 4n + 2}{n^2 + 3n + 1}\\ = \lim\dfrac{\dfrac{2}{\sqrt n} - \dfrac4n + \dfrac{2}{n^2}}{1 + \dfrac3n + \dfrac{1}{n^2}}\\ = \dfrac{0 - 0 + 0}{1 + 0 + 0}\\ = 0\\ g)\quad \lim\dfrac{n\sqrt n - n + 2}{n^2 + n\sqrt{3n} + 1}\\ = \lim \dfrac{\dfrac{1}{\sqrt n} - \dfrac1n + \dfrac{2}{n^2}}{1 + \sqrt{\dfrac3n} + \dfrac{1}{n^2}}\\ = \dfrac{0 - 0 + 0}{1 + 0 + 0}\\ =0\\ h)\quad \lim\dfrac{n^4 - 5n^2 + 4}{n^2 - 3n + 1}\\ = \lim\dfrac{n^2 - 5 + \dfrac{4}{n^2}}{1 - \dfrac3n + \dfrac{1}{n^2}}\\ = \dfrac{+\infty - 5 + 0}{1 - 0 + 0}\\ = +\infty\\ i)\quad \lim\dfrac{2n^3 - 3n + 3}{-n^2 + 3n - 2}\\ = \lim\dfrac{2n - \dfrac3n + \dfrac{3}{n^2}}{-1 + \dfrac3n - \dfrac{2}{n^2}}\\ = \dfrac{+\infty - 0 + 0}{-1 + 0 - 0}\\ = - \infty\\ j)\quad \lim\dfrac{n(n^2 + 2n)}{3 - 2n + 2n^2}\\ = \lim\dfrac{n\left(1 + \dfrac2n\right)}{\dfrac{3}{n^2} - \dfrac2n + 2}\\ = \dfrac{+\infty(1 + 0)}{0 - 0 + 2}\\ = +\infty \end{array}$
