Question

Tìm các giới hạn sau:

a) \(lim\sqrt[3]{-n^3+2n^2-5}\)

b) \(lim\dfrac{1}{\sqrt{n+1}-\sqrt{n}}\)

c) \(lim\left(\dfrac{1}{n+1}-n\right)\)

d) \(lim\left(\dfrac{2n^2-1}{n+1}-2n\right)\)

e) \(lim\dfrac{2n^3+n^2-3n+1}{2-3n}\)

Answer 1

\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)

\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)

\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)

\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)