Toán Tìm lim (n^3 + 4n – 5) / (3n^3 + n^2 + 7) 05/11/2021 By Mackenzie Tìm lim (n^3 + 4n – 5) / (3n^3 + n^2 + 7)
`\quad lim {n^3 + 4n – 5} / {3n^3 + n^2 + 7}` `=lim {n^3 (1+ 4/{n^2} – 5/{n^3})}/{n^3 (3+ 1/n + 7/{n^3})}` `=lim {1+4/{n^2} -5/{n^3}}/{3+1/ n + 7/ {n^3}}` `= {1+0+0}/{3+0+0}=1/ 3` Trả lời
$\,\,\,\,\lim \dfrac{{{n}^{3}}+4n-5}{3{{n}^{3}}+{{n}^{2}}+7}$ $=\lim \dfrac{1+\dfrac{4}{{{n}^{2}}}-\dfrac{5}{{{n}^{3}}}}{3+\dfrac{1}{n}+\dfrac{7}{{{n}^{3}}}}$ $=\dfrac{1}{3}$ Trả lời
`\quad lim {n^3 + 4n – 5} / {3n^3 + n^2 + 7}`
`=lim {n^3 (1+ 4/{n^2} – 5/{n^3})}/{n^3 (3+ 1/n + 7/{n^3})}`
`=lim {1+4/{n^2} -5/{n^3}}/{3+1/ n + 7/ {n^3}}`
`= {1+0+0}/{3+0+0}=1/ 3`
$\,\,\,\,\lim \dfrac{{{n}^{3}}+4n-5}{3{{n}^{3}}+{{n}^{2}}+7}$
$=\lim \dfrac{1+\dfrac{4}{{{n}^{2}}}-\dfrac{5}{{{n}^{3}}}}{3+\dfrac{1}{n}+\dfrac{7}{{{n}^{3}}}}$
$=\dfrac{1}{3}$