Tính \(L=\lim\dfrac{\left(n^2+2n\right)\left(2n^3+1\right)(4n+5)}{\left(n^4-3n-1\right)\left(3n^2-7\right)}\).
\(L=0\) | |
\(L=1\) | |
\(L=\dfrac{8}{3}\) | |
\(L=+\infty\) |
Chọn phương án C.
Dùng máy tính cầm tay:
Chọn phương án C.
\(\begin{aligned}
L&=\lim\dfrac{\left(n^2+2n\right)\left(2n^3+1\right)(4n+5)}{\left(n^4-3n-1\right)\left(3n^2-7\right)}\\
&=\lim\dfrac{n^2\left(1+\dfrac{2}{n}\right)n^3\left(2+\dfrac{1}{n^3}\right)n\left(4+\dfrac{5}{n}\right)}{n^4\left(1-\dfrac{3}{n^3}-\dfrac{1}{n^4}\right)n^2\left(3-\dfrac{7}{n^2}\right)}\\
&=\lim\dfrac{\left(1+\dfrac{2}{n}\right)\left(2+\dfrac{1}{n^3}\right)\left(4+\dfrac{5}{n}\right)}{\left(1-\dfrac{3}{n^3}-\dfrac{1}{n^4}\right)\left(3-\dfrac{7}{n^2}\right)}\\
&=\dfrac{\left(1+0\right)\left(2+0\right)\left(4+0\right)}{\left(1-0-0\right)\left(3-0\right)}\\
&=\dfrac{8}{3}.
\end{aligned}\)