Đáp án:
e) $\dfrac{1}{36}$
f) $\dfrac12$
Giải thích các bước giải:
e) $\lim\dfrac{(2n+1)(n-1)(3n-4)}{(6n-1)^3}$
$=\lim\dfrac{n\left(2+\dfrac1n\right).n\left(1-\dfrac1n\right).n\left(3 -\dfrac4n\right)}{n^3\left(6 -\dfrac1n\right)^3}$
$=\lim\dfrac{\left(2+\dfrac1n\right)\cdot\left(1-\dfrac1n\right)\cdot\left(3 -\dfrac4n\right)}{\left(6 -\dfrac1n\right)^3}$
$=\dfrac{(2+0)(1-0)(3-4.0)}{(6-0)^3}$
$=\dfrac{6}{6^3}$
$=\dfrac{1}{36}$
f) $\lim\dfrac{(3n+1)(n+2)(n+3)}{n(2n^2 + n+1)}$
$=\lim\dfrac{n\left(3 +\dfrac1n\right).n\left(1+\dfrac2n\right).n\left(1+\dfrac3n\right)}{n.n^2\left(2+\dfrac1n +\dfrac{1}{n^2}\right)}$
$=\lim\dfrac{\left(3 +\dfrac1n\right)\cdot\left(1+\dfrac2n\right)\cdot\left(1+\dfrac3n\right)}{2+\dfrac1n +\dfrac{1}{n^2}}$
$=\dfrac{(3+0)(1+2.0)(1+3.0)}{2+0+0}$
$=\dfrac32$
