Question

Tìm các giới hạn sau:

\(a,\left(8n-3n^9+1\right)\)

\(b,lim\left(6n^4-n+1\right)\)

\(c,lim\left(2-3n+7n^2\right)\)

Answer 1

\(a,lim\left(8n-3n^9+1\right)\)

\(=limn^9\left(\dfrac{8}{n^8}-3+\dfrac{1}{n^9}\right)\)

\(=n^9\left(0-3+0\right)=n^9.\left(-3\right)=\)-∞

 

Answer 2

\(\lim\left(6n^4-n+1\right)=\lim n^4\left(6-\dfrac{1}{n^3}+\dfrac{1}{n^4}\right)=+\infty.6=+\infty\)

\(\lim\left(2-3n+7n^2\right)=\lim n^2\left(\dfrac{2}{n^2}-\dfrac{3}{n}+7\right)=+\infty.7=+\infty\)